Thông tin tài liệu
❚❘×❮◆● ✣❸■ ❍➴❈ ❙× P❍❸▼ ❍⑨ ◆❐■ ✷
❑❍❖❆ ❚❖⑩◆
✯✯✯✯✯✯✯✯✯✯✯✯✯
❍❖⑨◆● ❚❍Ò❨ ▲■◆❍
P❍×❒◆● ❚❘➐◆❍
P❍×❒◆● P❍⑩P ●■❷■ ❱⑨ ▼❐❚ ❙➮ ❈⑩❈❍
❙⑩◆● ❚❸❖ ❘❆ ✣➋ ❚❖⑩◆ ▼❰■
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P ✣❸■ ❍➴❈
❈❤✉②➯♥ ♥❣➔♥❤✿ ✣↕✐ sè
◆❣÷í✐ ❤÷î♥❣ ❞➝♥ ❦❤♦❛ ❤å❝
❚❤❙✳ P❍❸▼ ▲×❒◆● ❇➀◆●
❍⑨ ◆❐■ ✲ ✷✵✶✺
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
▲❮■ ❈❷▼ ❒◆
❚r÷î❝ ❦❤✐ tr➻♥❤ ❜➔② ♥ë✐ ❞✉♥❣ ❝❤➼♥❤ ❝õ❛ ❦❤â❛ ❧✉➟♥ tèt ♥❣❤✐➺♣✱
❡♠ ①✐♥ ❜➔② tä ❧á♥❣ ❜✐➳t ì♥ s➙✉ s➢❝ tî✐ ❚❤↕❝ s➽ P❤↕♠ ▲÷ì♥❣ ❇➡♥❣✱
♥❣÷í✐ ✤➣ t➟♥ t➻♥❤ ❤÷î♥❣ ❞➝♥ ✤➸ ❡♠ ❝â t❤➸ ❤♦➔♥ t❤➔♥❤ ❦❤â❛ ❧✉➟♥ tèt
♥❣❤✐➺♣ ♥➔②✳
❊♠ ❝ô♥❣ ①✐♥ ❜➔② tä ❧á♥❣ ❜✐➳t ì♥ ❝❤➙♥ t❤➔♥❤ tî✐ t♦➔♥ t❤➸ t❤➛②✱
❝æ ❣✐→♦ tr♦♥❣ ❦❤♦❛ ❚♦→♥✱ ❚r÷í♥❣ ✣↕✐ ❤å❝ ❙÷ ♣❤↕♠ ❍➔ ◆ë✐ ✷ ✤➣ ❞↕②
❜↔♦ ❡♠ t➟♥ t➻♥❤ tr♦♥❣ s✉èt q✉→ tr➻♥❤ ❤å❝ t➟♣ t↕✐ ❦❤♦❛✳
◆❤➙♥ ❞à♣ ♥➔② ❡♠ ❝ô♥❣ ①✐♥ ✤÷ñ❝ ❣û✐ ❧í✐ ❝↔♠ ì♥ ❝❤➙♥ t❤➔♥❤ tî✐
❣✐❛ ✤➻♥❤✱ ❜↕♥ ❜➧ ✤➣ ❧✉æ♥ ❜➯♥ ❡♠✱ ✤ë♥❣ ✈✐➯♥✱ ❣✐ó♣ ✤ï ❡♠ tr♦♥❣ s✉èt
q✉→ tr➻♥❤ ❤å❝ t➟♣ ✈➔ t❤ü❝ ❤✐➺♥ ❦❤â❛ ❧✉➟♥ tèt ♥❣❤✐➺♣ ♥➔②✳
❳✉➙♥ ❍á❛✱ ♥❣➔② ✸✵ t❤→♥❣ ✹ ♥➠♠ ✷✵✶✺
❙✐♥❤ ✈✐➯♥
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✷
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
▲❮■ ❈❆▼ ✣❖❆◆
❊♠ ①✐♥ ❝❛♠ ✤♦❛♥ ✤➲ t➔✐ ♥➔② ❧➔ ❞♦ ❡♠ t❤ü❝ ❤✐➺♥✱ ✤â ❧➔ ❦➳t q✉↔
❝õ❛ q✉→ tr➻♥❤ ❤å❝ t➟♣ ✈➔ ♥❣❤✐➯♥ ❝ù✉ s→❝❤ ✈ð✱ t➔✐ ❧✐➺✉ ❝õ❛ ❡♠ ❞÷î✐ sü
❤÷î♥❣ ❞➝♥ ❝õ❛ ❚❤↕❝ s➽ P❤↕♠ ▲÷ì♥❣ ❇➡♥❣✱ ✤➲ t➔✐ ♥➔② ❦❤æ♥❣ trò♥❣
✈î✐ ❝→❝ ❦➳t q✉↔ tr÷î❝ ✤â ❝õ❛ ❝→❝ t→❝ ❣✐↔ ❦❤→❝✳
❳✉➙♥ ❍á❛✱ ♥❣➔② ✸✵ t❤→♥❣ ✹ ♥➠♠ ✷✵✶✺
❙✐♥❤ ✈✐➯♥
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✸
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
▼ö❝ ❧ö❝
■ ❑■➌◆ ❚❍Ù❈ ❈❒ ❇❷◆
✾
■■ ▼❐❚ ❙➮ P❍×❒◆● P❍⑩P ●■❷■ ❱⑨ ▼❐❚ ❙➮ ❈⑩❈❍ ❙⑩◆●
❚❸❖ ❘❆ ✣➋ ❚❖⑩◆ ▼❰■ ❱➋ P❍×❒◆● ❚❘➐◆❍
✶✷
✶ ▼ët sè ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
✶✳✶
✶✳✷
✶✳✸
✶✳✹
✶✳✺
✶✳✻
✶✳✼
✶✳✽
✶✳✾
P❤÷ì♥❣ ♣❤→♣ ❜✐➳♥ ✤ê✐ t÷ì♥❣ ✤÷ì♥❣ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
P❤÷ì♥❣ ♣❤→♣ ✤➦t ➞♥ ♣❤ö ✤÷❛ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ t➼❝❤
P❤÷ì♥❣ ♣❤→♣ sû ❞ö♥❣ ❤➺ sè ❜➜t ✤à♥❤ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
P❤÷ì♥❣ ♣❤→♣ ✤÷❛ ✈➲ ❤➺ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
P❤÷ì♥❣ ♣❤→♣ ❤➔♠ sè ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
P❤÷ì♥❣ ♣❤→♣ ❤➡♥❣ sè ❜✐➳♥ t❤✐➯♥ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
P❤÷ì♥❣ ♣❤→♣ sû ❞ö♥❣ ✤à♥❤ ❧þ ▲❛❣r❛♥❣❡ ✳ ✳ ✳ ✳ ✳ ✳ ✳
P❤÷ì♥❣ ♣❤→♣ ❤➻♥❤ ❤å❝ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
P❤÷ì♥❣ ♣❤→♣ ❜➜t ✤➥♥❣ t❤ù❝ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
✷ ▼ët sè ❝→❝❤ s→♥❣ t↕♦ r❛ ✤➲ t♦→♥ ♠î✐
✳
✳
✳
✳
✳
✳
✳
✳
✳
✳
✳
✳
✳
✳
✳
✳
✳
✳
✶✸
✶✸
✶✻
✶✾
✷✷
✷✼
✸✶
✸✹
✸✽
✸✾
✹✺
✷✳✶ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➺ sè ❜➜t
✤à♥❤ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✹✺
✷✳✷ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ✤÷❛ ✈➲ ❤➺ ✹✼
✷✳✸ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➔♠ sè ✳ ✺✶
✹
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✷✳✹ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➡♥❣ sè
❜✐➳♥ t❤✐➯♥ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
✷✳✺ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ①✉➜t ①ù tø ❤➻♥❤ ❤å❝ ✳ ✳ ✳ ✳ ✳ ✳ ✳
✷✳✻ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ sû ❞ö♥❣ ❜➜t
✤➥♥❣ t❤ù❝ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
✷✳✼ ▼ët sè ❤÷î♥❣ s→♥❣ t↕♦ ❝→❝ ♣❤÷ì♥❣ tr➻♥❤ ✤❛ t❤ù❝ ❜➟❝ ❝❛♦
✷✳✼✳✶ ❙û ❞ö♥❣ ♥❤ú♥❣ ♣❤÷ì♥❣ tr➻♥❤ ❝❤å♥ tr÷î❝ ✳ ✳ ✳ ✳
✷✳✼✳✷ ❙û ❞ö♥❣ ❝æ♥❣ t❤ù❝ ❧÷ñ♥❣ ❣✐→❝ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
✷✳✼✳✸ ❙û ❞ö♥❣ ♥❤à t❤ù❝ ◆✐✉✲tì♥ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
✷✳✽ ▼ët sè ❤÷î♥❣ s→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ♣❤➙♥ t❤ù❝ ❤ú✉ t➾
✷✳✾ ▼ët sè ❤÷î♥❣ s→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ ✳ ✳ ✳ ✳ ✳ ✳ ✳
✷✳✾✳✶ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ tø ❝→❝ ✤➥♥❣ t❤ù❝ ✳
✷✳✾✳✷ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ tø ❝→❝ ❤➺ ✤è✐ ①ù♥❣
❧♦↕✐ ■■ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
✷✳✾✳✸ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ ❞ü❛ ✈➔♦ t➼♥❤ ✤ì♥
✤✐➺✉ ❤➔♠ sè ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✺
✺✺
✺✾
✻✷
✻✾
✻✾
✼✵
✼✸
✼✺
✽✵
✽✵
✽✺
✽✽
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
▲❮■ ▼Ð ✣❺❯
✶✳ ▲➼ ❞♦ ❝❤å♥ ✤➲ t➔✐
❚r♦♥❣ ♥❤➔ tr÷í♥❣ ♣❤ê t❤æ♥❣✱ ♠æ♥ ❚♦→♥ ❣✐ú ♠ët ✈à tr➼ ❤➳t sù❝ q✉❛♥
trå♥❣✱ ❣✐ó♣ ❤å❝ s✐♥❤ r➧♥ ❧✉②➺♥ ✈➔ ❜ç✐ ❞÷ï♥❣ t÷ ❞✉② ❧æ❣✐❝✱ t➼♥❤ ❧✐♥❤
❤♦↕t✱ ❝➞♥ t❤➟♥✱ ❝ò♥❣ ✈î✐ ♥❤✐➲✉ ♥➠♥❣ ❧ü❝ tr➼ t✉➺ ❦❤→❝✳ ▼æ♥ ❚♦→♥ ❝á♥
❧➔ ❝æ♥❣ ❝ö ❝õ❛ ♥❤✐➲✉ ♥❣➔♥❤ ❦❤♦❛ ❤å❝ ❦➽ t❤✉➟t✱ ❝â ♥❤✐➲✉ ù♥❣ ❞ö♥❣ t♦
❧î♥ tr♦♥❣ ✤í✐ sè♥❣✳
❚r♦♥❣ ❝❤÷ì♥❣ tr➻♥❤ ♠æ♥ ❚♦→♥ ð ♥❤➔ tr÷í♥❣ ♣❤ê t❤æ♥❣✱ ♥ë✐ ❞✉♥❣
❞↕② ❤å❝ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ ❧➔ ♠ët ♥ë✐ ❞✉♥❣ q✉❛♥ trå♥❣✱ ♥â ❦❤æ♥❣ ♥❤ú♥❣
❧➔ ✤è✐ t÷ñ♥❣ ♥❣❤✐➯♥ ❝ù✉ ❝õ❛ ✣↕✐ sè ♠➔ ❝á♥ ❧➔ ❝æ♥❣ ❝ö ✤➢❝ ❧ü❝ ❝õ❛ ●✐↔✐
t➼❝❤✳
❚r♦♥❣ ❞↕② ❤å❝ t♦→♥✱ ♥❣♦➔✐ ✈✐➺❝ ✤↔♠ ❜↔♦ ❝✉♥❣ ❝➜♣ ✤➛② ✤õ ❝❤➼♥❤ ①→❝✱
❝â ❤➺ t❤è♥❣ ♥❤ú♥❣ ❦✐➳♥ t❤ù❝ ❝ì ❜↔♥ t❤➻ ♠ët ✤✐➲✉ q✉❛♥ trå♥❣ ❤ì♥ ❝↔ ❧➔
❧➔♠ s❛♦ ❤➻♥❤ t❤➔♥❤ ❝❤♦ ❤å❝ s✐♥❤ ♣❤÷ì♥❣ ♣❤→♣ ❝❤✉♥❣ ✤➸ ❣✐↔✐ ❝→❝ ❞↕♥❣
t♦→♥✱ tø ✤â ❣✐ó♣ ❝→❝ ❡♠ t➼❝❤ ❝ü❝ ❤♦↕t ✤ë♥❣✱ ✤ë❝ ❧➟♣ s→♥❣ t↕♦ ✤➸ ❞➛♥
❤♦➔♥ t❤✐➺♥ ❦➽ ♥➠♥❣✱ ❦➽ ①↔♦✱ ♣❤→t tr✐➸♥ ♥➠♥❣ ❧ü❝ t÷ ❞✉② ✈➔ ❤♦➔♥ t❤✐➺♥
♥❤➙♥ ❝→❝❤✳
✣è✐ ✈î✐ ♥ë✐ ❞✉♥❣ ♣❤÷ì♥❣ tr➻♥❤ tr♦♥❣ ❝❤÷ì♥❣ tr➻♥❤ t♦→♥ ♣❤ê t❤æ♥❣✱
❝â r➜t ♥❤✐➲✉ ♥❤ú♥❣ ❜➔✐ t♦→♥ ✤❛ ❞↕♥❣✳ ◆❣÷í✐ ❣✐→♦ ✈✐➯♥ ❝➛♥ ♣❤↔✐ ♥➢♠
❜➢t ✤÷ñ❝ ♥❤ú♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐✱ ♥❤ú♥❣ ❤÷î♥❣ ✤✐ ♣❤♦♥❣ ♣❤ó ✤➸ tø
✤â ①→❝ ✤à♥❤ ❝❤♦ ❤å❝ s✐♥❤ ❝♦♥ ✤÷í♥❣ ♥➔♦ ❣✐↔✐ q✉②➳t ❜➔✐ t♦→♥ ♥❤❛♥❤
❝❤â♥❣ ✈➔ ❝❤➼♥❤ ①→❝ ♥❤➜t✱ ❦❤æ♥❣ ♥❤ú♥❣ t❤➳ ❣✐→♦ ✈✐➯♥ ❝á♥ ♣❤↔✐ ❜✐➳t ①➙②
❞ü♥❣ s→♥❣ t↕♦ ❝→❝ ✤➲ t♦→♥ ✤➸ ❧➔♠ t➔✐ ❧✐➺✉ ❝❤♦ ✈✐➺❝ ❣✐↔♥❣ ❞↕②✱ ✈î✐ ♠ö❝
✤➼❝❤ ✤÷❛ ✤➳♥ ❝❤♦ ❤å❝ s✐♥❤ ♥❤ú♥❣ ✤➲ t♦→♥ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ ❤❛②✱ ♠î✐
♠➫✱ ♥❤➡♠ ❝õ♥❣ ❝è r➧♥ ❧✉②➺♥ ✤ó♥❣ ♥ë✐ ❞✉♥❣ ❦✐➳♥ t❤ù❝ ❝➛♥ ♥➢♠✳
❱î✐ ♥❤ú♥❣ ❧➼ ❞♦ tr➯♥ ❝ò♥❣ ✈î✐ ❧á♥❣ s❛② ♠➯ ♥❣❤✐➯♥ ❝ù✉ ✈➔ ✤÷ñ❝ sü
❣✐ó♣ ✤ï t➟♥ t➻♥❤ ❝õ❛ ❚❤↕❝ s➽ P❤↕♠ ▲÷ì♥❣ ❇➡♥❣✱ ❡♠ ✤➣ ❝❤å♥ ✤➲ t➔✐✿
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✻
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✧P❤÷ì♥❣ tr➻♥❤ ✲ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ✈➔ ♠ët sè ❝→❝❤ s→♥❣ t↕♦ r❛
✤➲ t♦→♥ ♠î✐✧ ✤➸ ❧➔♠ ❦❤â❛ ❧✉➟♥ tèt ♥❣❤✐➺♣ ✈î✐ ♠♦♥❣ ♠✉è♥ ❣â♣ ♣❤➛♥
❜➨ ♥❤ä ❧➔♠ t➠♥❣ ✈➫ ✤➭♣ ❝õ❛ ♠æ♥ ❚♦→♥ q✉❛ ♥ë✐ ❞✉♥❣ ♣❤÷ì♥❣ tr➻♥❤✳
✷✳ ▼ö❝ ✤➼❝❤ ♥❣❤✐➯♥ ❝ù✉
❇÷î❝ ✤➛✉ ❧➔♠ q✉❡♥ ✈î✐ ♥❣❤✐➯♥ ❝ù✉ ❦❤♦❛ ❤å❝ ✈➔ t➻♠ ❤✐➸✉ s➙✉ ❤ì♥
✈➲ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ ✈➔ ❝→❝❤ t↕♦ r❛ ♥❤ú♥❣ ✤➲ t♦→♥ ♠î✐✳
✸✳ ✣è✐ t÷ñ♥❣ ✈➔ ♣❤↕♠ ✈✐ ♥❣❤✐➯♥ ❝ù✉
✲ ✣è✐ t÷ñ♥❣ ♥❣❤✐➯♥ ❝ù✉✿ ♣❤÷ì♥❣ tr➻♥❤
✲ P❤↕♠ ✈✐ ♥❣❤✐➯♥ ❝ù✉✿ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾✱ ♣❤÷ì♥❣ tr➻♥❤ ✤❛ t❤ù❝✱
♣❤÷ì♥❣ tr➻♥❤ ♣❤➙♥ t❤ù❝ ❤ú✉ t➾✱ ♣❤÷ì♥❣ tr➻♥❤ ♠ô ✈➔ ❧æ❣❛r✐t✳
✹✳ ◆❤✐➺♠ ✈ö ♥❣❤✐➯♥ ❝ù✉
✲ ❑✐➳♥ t❤ù❝ ❝ì ❜↔♥ ✈➲ ♣❤÷ì♥❣ tr➻♥❤
✲ ❈→❝ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
✲ ❈→❝ ♣❤÷ì♥❣ ♣❤→♣ s→♥❣ t↕♦ r❛ ✤➲ t♦→♥ ♠î✐ ✈➲ ♣❤÷ì♥❣ tr➻♥❤
✺✳ P❤÷ì♥❣ ♣❤→♣ ♥❣❤✐➯♥ ❝ù✉
✲ P❤÷ì♥❣ ♣❤→♣ ♥❣❤✐➯♥ ❝ù✉ t➔✐ ❧✐➺✉
✲ P❤÷ì♥❣ ♣❤→♣ s♦ s→♥❤✱ ♣❤➙♥ t➼❝❤✱ tê♥❣ ❤ñ♣
✲ P❤÷ì♥❣ ♣❤→♣ ✤→♥❤ ❣✐→
✻✳ ❈➜✉ tró❝ ❦❤â❛ ❧✉➟♥
✲ ◆ë✐ ❞✉♥❣ ❦❤â❛ ❧✉➟♥ ❝õ❛ ❡♠ ❣ç♠✿
P❤➛♥ ✶✿ ❑✐➳♥ t❤ù❝ ❝ì ❜↔♥
P❤➛♥ ✷✿ ▼ët sè ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ✈➔ ♠ët sè ❝→❝❤ s→♥❣ t↕♦ r❛
✤➲ t♦→♥ ♠î✐ ✈➲ ♣❤÷ì♥❣ tr➻♥❤
❈❤÷ì♥❣ ✶✳ ▼ët sè ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
❈❤÷ì♥❣ ✷✳ ▼ët sè ❝→❝❤ s→♥❣ t↕♦ r❛ ✤➲ t♦→♥ ♠î✐
✲ P❤➛♥ ✶✱ ❡♠ tr➻♥❤ ❜➔② ❧↕✐ ♥❤ú♥❣ ❦✐➳♥ t❤ù❝ ❝ì ❜↔♥ ✈➲ ♣❤÷ì♥❣ tr➻♥❤
❜❛♦ ❣ç♠✿ ❦❤→✐ ♥✐➺♠ ♣❤÷ì♥❣ tr➻♥❤✱ t➟♣ ①→❝ ✤à♥❤✱ ♥❣❤✐➺♠✱ t➟♣ ♥❣❤✐➺♠✱
❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ ❧➔ ❣➻✱ ♣❤÷ì♥❣ tr➻♥❤ t÷ì♥❣ ✤÷ì♥❣ ✈➔ ♣❤÷ì♥❣ tr➻♥❤ ❤➺
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✼
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
q✉↔✳
✲ P❤➛♥ ✷✱ ❧➔ ♥ë✐ ❞✉♥❣ ❝❤➼♥❤ ❝õ❛ ❦❤â❛ ❧✉➟♥ ❝❤✐❛ ❧➔♠ ✷ ❝❤÷ì♥❣✿
❚r♦♥❣ ❝❤÷ì♥❣ ✶✱ ❡♠ tr➻♥❤ ❜➔② ✾ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤✱
tr♦♥❣ ✤â ❝â ❝→❝ ♣❤÷ì♥❣ ♣❤→♣ r➜t q✉❡♥ t❤✉ë❝ ♥❤÷✿ ❜✐➳♥ ✤ê✐ t÷ì♥❣
✤÷ì♥❣✱ ✤➦t ➞♥ ♣❤ö ✤÷❛ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ t➼❝❤✱✳✳✳❱➔ ♠ët ✈➔✐ ♣❤÷ì♥❣
♣❤→♣ ❤❛②✱ ❦❤→ ♠î✐ ♠➫ tr♦♥❣ t❤í✐ ❣✐❛♥ ❣➛♥ ✤➙② ♥❤÷ ♣❤÷ì♥❣ ♣❤→♣ sû
❞ö♥❣ ✤à♥❤ ❧þ ▲❛❣r❛♥❣❡✱ ♣❤÷ì♥❣ ♣❤→♣ ❤➻♥❤ ❤å❝✱ ♣❤÷ì♥❣ ♣❤→♣ ❤➡♥❣ sè
❜✐➳♥ t❤✐➯♥✳✳✳
❚r♦♥❣ ❝❤÷ì♥❣ ✷✱ ❡♠ tr➻♥❤ ❜➔② ❝→❝ ♣❤÷ì♥❣ ♣❤→♣ s→♥❣ t↕♦ r❛
❜➔✐ t♦→♥ ♠î✐ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ tr➯♥ ❝ì sð ①✉➜t ♣❤→t tø ♣❤÷ì♥❣ ♣❤→♣
❣✐↔✐✳ ❱➼ ❞ö✱ s→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤✿ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ✤÷❛ ✈➲ ❤➺✱
❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➔♠ sè✱✳✳✳P❤➛♥ t✐➳♣ t❤❡♦✱ ❡♠ tr➻♥❤ ❜➔② ♠ët sè
♣❤÷ì♥❣ ♣❤→♣ s→♥❣ t↕♦ r❛ ♣❤÷ì♥❣ tr➻♥❤ ✤❛ t❤ù❝ ❜➟❝ ❝❛♦✱ ♥❤÷ ♣❤÷ì♥❣
♣❤→♣✿ sû ❞ö♥❣ ♥❤ú♥❣ ♣❤÷ì♥❣ tr➻♥❤ ❝❤å♥ tr÷î❝✱ sû ❞ö♥❣ ❝æ♥❣ t❤ù❝
❧÷ñ♥❣ ❣✐→❝✱ sû ❞ö♥❣ ♥❤à t❤ù❝ ◆✐✉✲tì♥✳ ❙❛✉ ✤â ❧➔ ♠ö❝ s→♥❣ t↕♦ ♣❤÷ì♥❣
tr➻♥❤ ♣❤➙♥ t❤ù❝ ❤ú✉ t➾✳ ❈✉è✐ ❝ò♥❣✱ ❡♠ tr➻♥❤ ❜➔② ♠ët sè ❤÷î♥❣ s→♥❣
t↕♦ r❛ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾✿ s→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ tø ❝→❝ ✤➥♥❣
t❤ù❝✱ s→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ tø ❝→❝ ❤➺ ✤è✐ ①ù♥❣ ❧♦↕✐ ■■✱ s→♥❣ t↕♦
♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ ❞ü❛ ✈➔♦ t➼♥❤ ✤ì♥ ✤✐➺✉ ❤➔♠ sè✳
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✽
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
P❤➛♥ ■
❑■➌◆ ❚❍Ù❈ ❈❒ ❇❷◆
✾
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
P❍❺◆ ✶✳
❑■➌◆ ❚❍Ù❈ ❈❒ ❇❷◆
✶✳ ❑❤→✐ ♥✐➺♠ ♣❤÷ì♥❣ tr➻♥❤
❛✮ ✣à♥❤ ♥❣❤➽❛
❈❤♦ ❤➔♠ sè f (x) ①→❝ ✤à♥❤ tr➯♥ D1
D2 ⊂ Rn ✈î✐ x = (x1 , x2 , ..., xn ) ∈ Rn
●å✐ M ❧➔ t➟♣ t➜t ❝↔ ❝→❝ ♠➺♥❤ ✤➲✳
❚❛ ❣å✐ ❤➔♠ ♠➺♥❤ ✤➲✿
⊂ Rn
✈➔ g(x) ①→❝ ✤à♥❤ tr➯♥
D = D1 ∩ D2 → M
x → ✧f (x) = g (x) ✧
❧➔ ♠ët ♣❤÷ì♥❣ tr➻♥❤ n ➞♥✳ ❚➟♣ D ✤÷ñ❝ ❣å✐ ❧➔ t➟♣ ①→❝ ✤à♥❤ ❝õ❛ ♣❤÷ì♥❣
tr➻♥❤✳
❚❛ t❤÷í♥❣ ✈✐➳t ♣❤÷ì♥❣ tr➻♥❤ ❞÷î✐ ❞↕♥❣ f (x) = g(x) ✈➔ ❤✐➸✉ t➟♣ ①→❝
✤à♥❤ D ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ♥➔② ✿ D = x ∈ Rn|f (x), g(x) tç♥ t↕✐ ✈➔
♥❤ú♥❣ ✤✐➲✉ ❦✐➺♥ ❝õ❛ x ♠➔ ♣❤÷ì♥❣ tr➻♥❤ ②➯✉ ❝➛✉✳
✣➦❝ ❜✐➺t✱ ♥➳✉ n = 1 t❤➻ ♣❤÷ì♥❣ tr➻♥❤ f (x) = g (x) ✤÷ñ❝ ❣å✐ ❧➔
♣❤÷ì♥❣ tr➻♥❤ ♠ët ➞♥✳
❜✮ ✣à♥❤ ♥❣❤➽❛
❈❤♦ ♣❤÷ì♥❣ tr➻♥❤ f (x) = g (x) ❝â t➟♣ ①→❝ ✤à♥❤ ❧➔ D✱ ♥➳✉ ❝â a ∈ D
s❛♦ ❝❤♦ ♠➺♥❤ ✤➲ ✧f (a) = g (a)✧ ✤ó♥❣ t❤➻ a ✤÷ñ❝ ❣å✐ ❧➔ ♠ët ♥❣❤✐➺♠
❝õ❛ ♣❤÷ì♥❣ tr➻♥❤✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ ❧➔ ✈✐➺❝ t➻♠ t➟♣ ♥❣❤✐➺♠ S ❝õ❛ ♥â✳
◆➳✉ S = φ t❤➻ ♣❤÷ì♥❣ tr➻♥❤ ✈æ ♥❣❤✐➺♠✳
✷✳ P❤÷ì♥❣ tr➻♥❤ t÷ì♥❣ ✤÷ì♥❣ ✈➔ ♣❤÷ì♥❣ tr➻♥❤ ❤➺ q✉↔
❛✮ ✣à♥❤ ♥❣❤➽❛
❈❤♦ ❤❛✐ ♣❤÷ì♥❣ tr➻♥❤
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✶✵
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✭✶✮
f1 (x) = g1 (x) ✭✷✮
− ◆➳✉ t➟♣ ♥❣❤✐➺♠ ❝õ❛ ❤❛✐ ♣❤÷ì♥❣ tr➻♥❤ ♥➔② ❜➡♥❣ ♥❤❛✉ t❤➻ t❛ ♥â✐
❤❛✐ ♣❤÷ì♥❣ tr➻♥❤ ❧➔ t÷ì♥❣ ✤÷ì♥❣✳ ❑➼ ❤✐➺✉✿ (1) ⇔ (2)✳
− ◆➳✉ ♠å✐ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✭✶✮ ✤➲✉ ❧➔ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣
tr➻♥❤ ✭✷✮ t❤➻ ♣❤÷ì♥❣ tr➻♥❤ ✭✷✮ ❣å✐ ❧➔ ♣❤÷ì♥❣ tr➻♥❤ ❤➺ q✉↔ ❝õ❛ ♣❤÷ì♥❣
tr➻♥❤ ✭✶✮✳ ❑➼ ❤✐➺✉✿ (1) ⇒ (2)✳
❈❤ó þ✿ ❍❛✐ ♣❤÷ì♥❣ tr➻♥❤ ✈æ ♥❣❤✐➺♠ ❧✉æ♥ t÷ì♥❣ ✤÷ì♥❣✳
f (x) = g (x)
❜✮ ▼ët sè ♣❤➨♣ ❜✐➳♥ ✤ê✐ t÷ì♥❣ ✤÷ì♥❣ t❤÷í♥❣ ❣➦♣
❈❤♦ ♣❤÷ì♥❣ tr➻♥❤ f (x) = g (x) ✭✶✮ ❝â t➟♣ ①→❝ ✤à♥❤ ❧➔ D ⊂ Rn
• ◆➳✉ ❤➔♠ sè h(x) ①→❝ ✤à♥❤ tr➯♥ D t❤➻ ♣❤÷ì♥❣ tr➻♥❤
(1) ⇔ f (x) + h (x) = g (x) + h (x)
•
◆➳✉ ❤➔♠ sè h(x) ①→❝ ✤à♥❤ ✈➔ ❦❤→❝ ✵ tr➯♥ D t❤➻
(1) ⇔ f (x) .h (x) = g (x) .h (x)
•
❚❛ ❝â✿
✰ (1) ⇔ (f (x))2n+1 = (g (x))2n+1, ∀n ∈ N
✰ (1) ⇒ (f (x))2n = (g (x))2n, ∀n ∈ N∗
✰ ◆➳✉ f (x) ✈➔ g(x) ❝ò♥❣ ❞÷ì♥❣ ❤♦➦❝ ❝ò♥❣ ➙♠ tr➯♥ D t❤➻
(1) ⇔ (f (x))2n = (g (x))2n , ∀n ∈ N∗
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✶✶
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
P❤➛♥ ■■
▼❐❚ ❙➮ P❍×❒◆● P❍⑩P ●■❷■
❱⑨ ▼❐❚ ❙➮ ❈⑩❈❍ ❙⑩◆● ❚❸❖
❘❆ ✣➋ ❚❖⑩◆ ▼❰■ ❱➋
P❍×❒◆● ❚❘➐◆❍
✶✷
❈❤÷ì♥❣ ✶
▼ët sè ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ♣❤÷ì♥❣
tr➻♥❤
✶✳✶ P❤÷ì♥❣ ♣❤→♣ ❜✐➳♥ ✤ê✐ t÷ì♥❣ ✤÷ì♥❣
❈❤ó♥❣ t❛ sû ❞ö♥❣ ♠ët sè ♣❤➨♣ ❜✐➳♥ ✤ê✐ ❝ì ❜↔♥ s❛✉✿
✲ ✣è✐ ✈î✐ ♣❤÷ì♥❣ tr➻♥❤ ❝❤ù❛ ❝➠♥ t❤ù❝✿
• f (x) = g (x) ⇔ f (x) = g (x) ≥ 0 ✭ ✈î✐ ✤✐➲✉ ❦✐➺♥ f (x) , g (x) ❝â
♥❣❤➽❛✮✳
g (x) ❝â ♥❣❤➽❛ ✈➔ g (x) ≥ 0
• f (x) = g (x) ⇔
f (x) = g 2 (x)
✭❦❤æ♥❣ ❝➛♥ ✤✐➲✉ ❦✐➺♥ f (x) ≥ 0✮✳
f (x) ≥ 0
• f (x)+ g (x) = h (x) ⇔
g (x) ≥ 0
f (x) + g (x) + 2
f (x) g (x) = h (x) .
❝➛♥ h (x) ≥ 0✮✳
✭✈î✐ ✤✐➲✉ ❦✐➺♥ f (x) , g (x) , h (x) ❝â ♥❣❤➽❛ ✈➔ ❦❤æ♥❣
✲ ✣è✐ ✈î✐ ♣❤÷ì♥❣ tr➻♥❤ ❧♦❣❛r✐t✿ ✤➸ ❝❤✉②➸♥ ➞♥ sè r❛ ❦❤ä✐ ❧♦❣❛ ♥❣÷í✐
t❛ ❝â t❤➸ ♠ô ❤â❛ t❤❡♦ ❝ò♥❣ ♠ët ❝ì sè ❝↔ ❤❛✐ ✈➳ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤✳
❈❤ó♥❣ t❛ ❧÷✉ þ ❝→❝ ♣❤➨♣ ❜✐➳♥ ✤ê✐ s❛✉✿
• loga f (x) = loga g (x) ⇔
0 0
✶✸
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
• loga f (x) = b ⇔ f (x) = ab ✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ x − √2x + 3 = 0
●✐↔✐✳ P❤÷ì♥❣ tr➻♥❤ ✈✐➳t ❧↕✐ ❞÷î✐ ❞↕♥❣✿
√
x≥0
2x + 3 = x ⇔
⇔
2
2x + 3 = x
x≥0
⇔
x = −1 ⇔ x = 3.
x=3
x≥0
x2 − 2x − 3 = 0
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ x = 3✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ −4 ≤ x ≤ 21 .
√
x+4−
√
1−x=
√
1 − 2x.
P❤÷ì♥❣ tr➻♥❤ ✈✐➳t ❧↕✐ ❞÷î✐ ❞↕♥❣✿
√
❱➟②
1−x+
√
1 − 2x =
√
x+4⇔
(1 − x) (1 − 2x) = 2x + 1
x ≥ −1
2x + 1 ≥ 0
2
⇔
⇔
2x2 + 7x = 0
(1 − x) (1 − 2x) = (2x + 1)2
1
x
≥
−
2
⇔x=0
⇔
x=0
x = −7
2
♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ x = 0✳
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥
x2 + 4x − 4 > 0
0 −2 +
√
6
√
x < −2 − 6
⇔
√
6 − 2 < x = 1.
00
x+6>0
−6 < x < 4, x = −2
⇔
⇔
x+2=0
4 |x + 2| = −x2 − 2x + 24
4 |x + 2| = (4 − x) (x + 6)
−6 < x < −2
−6 < x < −2
−4x − 8 = −x2 − 2x + 24
x2 − 2x − 32 = 0
⇔
⇔
−2 < x < 4
−2 < x < 4
2
4x + 8 = −x − 2x + 24
x2 + 6x − 16 = 0
−6 < x < −2
√
x = 1 ± 33
√
x
=
1
−
33
⇔
⇔
−2
<
x
<
4
x = 2.
x=2
x = −8
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✶✺
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❱➟② ✭✶✮ ❝â ❤❛✐ ♥❣❤✐➺♠ x = 1 −
√
33
✈➔ x = 2✳
✶✳✷ P❤÷ì♥❣ ♣❤→♣ ✤➦t ➞♥ ♣❤ö ✤÷❛ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ t➼❝❤
❳✉➜t ♣❤→t tø ♠ët sè ❤➡♥❣ ✤➥♥❣ t❤ù❝ ❝ì ❜↔♥ ❦❤✐ ✤➦t ➞♥ ♣❤ö✿
• x3 + 1 = (x + 1) x2 − x + 1
√
√
• x4 + 1 = x2 − 2x + 1 x2 + 2x + 1
• x4 + x2 + 1 = x4 + 2x2 + 1 − x2 = x2 + x + 1
• 4x4 + 1 = 2x2 − 2x + 1
x2 − x + 1
2x2 + 2x + 1
▼ö❝ ✤➼❝❤ ❝õ❛ t❛ s❛✉ ❦❤✐ ✤➦t ➞♥ ♣❤ö ❧➔ ✤÷❛ ✈➲ ♥❤ú♥❣ ❞↕♥❣ ❝ì ❜↔♥ ✈➼
❞ö ♥❤÷✿
u + v = 1 + uv ⇔ (u − 1) (v − 1) = 0
❤❛②
au + bv = ab + vu ⇔ (u − b) (v − a) = 0
◆❣♦➔✐ r❛ ❝á♥ ♥❤✐➲✉ ❞↕♥❣ ❦❤→❝ ♥ú❛✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
√
√
2 1 − x − 1 + x + 3 1 − x2 = 3 − x✳
●ñ✐ þ✳ ❈❤å♥√α, β s❛♦ ❝❤♦ √
−x + 3 = α
1−x
2
+β
1+x
2
⇔ −x + 3 = (−α + β) x + α + β ⇒
−α + β = −1
α+β =3
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ −1 ≤ x ≤ 1✳ P❤÷ì♥❣ tr➻♥❤ ✈✐➳t ❧↕✐
⇔
α=2
β = 1.
√
√
√
(1 + x) + 2 (1 − x) − 2 1 − x + 1 + x − 3 1 − x2 = 0
√
√
✣➦t u = 1 + x ≥ 0, v = 1 − x ≥ 0. ❚❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝
✭✶✮
u2 + 2v 2 − 2v + u − 3uv = 0
⇔ (u − 2v) (u − v + 1) = 0
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✶✻
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
⇔
u = 2v
u=v−1
• ❑❤✐ u = 2v ✱ t❛ ❝â
√
√
3
1 + x = 2 1 − x ⇔ 1 + x = 4 − 4x ⇔ 5x = 3 ⇔ x = ✳
5
• ❑❤✐ v = u + 1✱ t❛ ❝â
√
√
√
1+x+1= 1−x⇔2 1+x+2+x=1−x
√
√
−2x − 1 > 0
3
⇔x=− .
⇔ 2 1 + x = −2x − 2 ⇔
2
4 + 4x = 4x2 + 4x + 1
❑➳t ❤ñ♣ ✈î✐ ✤✐➲✉ ❦✐➺♥ ✤÷ñ❝ t➟♣ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❧➔
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
1+x−1
√
√
3
3
,−
5
2
✳
1 − x + 1 = 2x
●✐↔✐✳ √✣✐➲✉ ❦✐➺♥✿ −1 ≤ x ≤ √1 ✭✶✮
✣➦t
t❤➔♥❤
1+x=u 0≤u≤
2
✳ ❙✉② r❛ x = u2 − 1✱ ♣❤÷ì♥❣ tr➻♥❤ trð
√
2 − u2 + 1 = 2 u2 − 1
√
⇔ (u − 1)
2 − u2 + 1 − 2 (u + 1) = 0
√
⇔ (u − 1) 2 − u2 − 2u − 1 = 0
(u − 1)
u−1=0
√
2 − u2 − 2u − 1 = 0
• ◆➳✉ u − 1 = 0 ⇒ u = 1 ✭t❤ä❛
⇔
♠➣♥ u ≥ 0✮ ✱ s✉② r❛ x = 0 t❤ä❛ ♠➣♥
✭✶✮
√
√
• ◆➳✉ 2 − u2 − 2u − 1 = 0 ⇔ 2 − u2 = 2u + 1
⇔
2u + 1 ≥ 0
2 − u2 = (2u + 1)2
−1
u
=
−1
<
2
⇔ 5u2 + 4u − 1 = 0 ⇔
1
u=
5
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✶✼
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❱➟② t❛ ❝â x = u
2
−1=
2
1
5
−1=
−24
25
✭t❤ä❛ ♠➣♥ ✤✐➲✉ ❦✐➺♥ ✭✶✮✮
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ 2x −5x+6 + 21−x
●✐↔✐✳ ❱✐➳t ❧↕✐ ♣❤÷ì♥❣ tr➻♥❤ ❞÷î✐ ❞↕♥❣✿
2
2x
2
2
= 2.26−5x + 1.
2
−5x+6
+ 21−x = 27−5x + 1
2
2
2
2
⇔ 2x −5x+6 + 21−x = 2(x −5x+6)+(1−x ) + 1
⇔ 2x
2
−5x+6
✣➦t
2
+ 21−x = 2x
u = 2x
2
2
−5x+6 1−x2
2
+1
−5x+6
v = 21−x
, (u, v > 0).
2
❑❤✐ ✤â✱ ♣❤÷ì♥❣ tr➻♥❤ t÷ì♥❣ ✤÷ì♥❣ ✈î✐✿
u + v = uv + 1 ⇔ (u − 1) (v − 1) = 0
⇔
u=1
v=1
⇔
2x
2
−5x+6
=1
2
21−x = 1
x2 − 5x + 6 = 0
⇔
1 − x2 = 0
x=3
⇔
x=2
x = ±1.
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ❜è♥ ♥❣❤✐➺♠ x = 3, x = 2, x = ±1.
❇➔✐ t♦→♥ ✹✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ √x + 1 + √x + 2 = 1 + √x2 + 3x + 2
❍÷î♥❣ ❞➝♥✳ ❚❛ t❤➜② (x + 1) (x + 2) = x2 + 3x + 2 ♥➯♥ ✤➦t
u=
√
3
3
3
3
x + 1, v =
√
3
x+2
✤➸ ❝â ♣❤÷ì♥❣ tr➻♥❤ u + v = 1 + uv ⇔ (u − 1) (v − 1) = 0. ❚ø ✤â t➻♠
✤÷ñ❝ ♥❣❤✐➺♠ x ∈ {0; −1}
❇➔✐ t♦→♥ ✺✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
1+
2
3
x − x2 =
Þ t÷ð♥❣✳ ❚❛ t❤➜② (√x)2 +
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
√
√
x+
1−x
✶✽
√
2
1−x
=1
(1)
✭✷✮✱ ♠➔ tø ♣❤÷ì♥❣ tr➻♥❤
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✤➛✉ t❛ rót ✤÷ñ❝ ♠ët ❝➠♥ t❤ù❝ q✉❛ ❝➠♥ t❤ù❝ ❝á♥ ❧↕✐✳ ❱➟② t❛ ❝â ❤÷î♥❣
❣✐↔✐ ♥❤÷ s❛✉✳
√
√
3 1−x−3
❍÷î♥❣ ❞➝♥✳ ❳➨t ✭✶✮ t❛ ❝â (1) ⇔ x = √
2 1−x−3
3t − 3
❉♦ ✤â ✤➦t t = 1 − x ⇒ x = 2t − 3 ❚❤❛② ✈➔♦ ✭✷✮ t❛ ❜✐➳♥ ✤ê✐ t❤➔♥❤
t (t − 1) 2t2 − 4t + 3 = 0 ⇔ t ∈ {0; 1}
√
√
❚ø ✤â s✉② r❛ x ∈ {0; 1}✳
✶✳✸ P❤÷ì♥❣ ♣❤→♣ sû ❞ö♥❣ ❤➺ sè ❜➜t ✤à♥❤
P❤÷ì♥❣ ♣❤→♣ ❤➺ sè ❜➜t ✤à♥❤ ❧➔ ❝❤➻❛ ❦❤â❛ ❣✐ó♣ t❛ ♣❤➙♥ t➼❝❤✱ t➻♠
✤÷ñ❝ ❧í✐ ❣✐↔✐ ❝❤♦ ♥❤✐➲✉ ❧♦↕✐ ♣❤÷ì♥❣ tr➻♥❤✳ ❚r♦♥❣ ♣❤➛♥ ♥➔②✱ t❛ s➩ t➻♠
❤✐➸✉ ♣❤÷ì♥❣ ♣❤→♣ ♥➔② t❤æ♥❣ q✉❛ ♠ët sè ❜➔✐ t♦→♥ s❛✉ ✤➙②✱ ð ✤â ❝â
♣❤➛♥ tr➻♥❤ ❜➔② þ t÷ð♥❣ ✈➔ ❝→❝ ❣ñ✐ þ ❝→❝❤ ❧➔♠✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣√tr➻♥❤✿
5 x(2x + 1) +
√
2x + 1 − 3 x = 8x + 1
✭✶✮
Þ t÷ð♥❣✳ P❤÷ì♥❣ tr➻♥❤ ✭✶✮ ❦❤✐➳♥ t❛ ❜è✐ rè✐ ✈➻ ♥➳✉ t❤ü❝ ❤✐➺♥ ♣❤➨♣
❧ô② t❤ø❛ ✤➸ ❣✐↔♠ ❜ît ❝➠♥ t❤ù❝ t❤➻ s➩ ❣➦♣ ♥❤✐➲✉ ❦❤â ❦❤➠♥✳ ▼ët s✉②
♥❣❤➽ tü ♥❤✐➯♥ ❧➔ t❛ s➩ ❜✐➸✉ ❞✐➵♥ 8x + 1 t❤❡♦ ❤❛✐ ❜✐➸✉ t❤ù❝ ♥➡♠ tr♦♥❣
❝➠♥ ❧➔ 2x + 1 ✈➔ x✳ ❱➔ ❝→❝❤ tèt ♥❤➜t ❧➔ ❤➺ sè ❜➜t ✤à♥❤✿
❚❛ t➻♠ α✱ β s❛♦ ❝❤♦
8x + 1 = α (2x + 1) + βx ⇔
2α + β = 8
α=1
⇔
α=1
β=6
❚❛ ❝â ❧í✐ ❣✐↔✐✳
✣✐➲✉ ❦✐➺♥✿ x ≥ 0✳ ❑❤✐ ✤â✿
(1) ⇔ 5 x (2x + 1) +
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
√
√
2x + 1 − 3 x = (2x + 1) + 6x
✶✾
✭✷✮
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✣➦t u =
√
2x + 1✱v =
√
x✳
❚❤❛② ✈➔♦ ✭✷✮ t❛ ✤÷ñ❝✿
5uv + u − 3v = u2 + 6v 2
⇔ (u − 3v) + 3uv − u2 + 2uv − 6v 2 = 0
⇔ (u − 3v) − u (u − 3v) + 2v (u − 3v) = 0
⇔
⇔ (u − 3v) (1 − u + 2v) = 0
√
√
u − 3v = 0
2x + 1 − 3 x = 0
⇔
√
√
1 − u + 2v = 0
1 − 2x + 1 + 2 x = 0
√
●✐↔✐ ✭✸✮✿ (3) ⇔
✭t❤ä❛ ✤✐➲✉ ❦✐➺♥✮
●✐↔✐ ✭✹✮✿
(3)
(4)
√
1
2x + 1 = 3 x ⇔ 2x + 1 = 9x ⇔ x =
7
√
√
√
(4) ⇔ 1 + 2 x = 2x + 1 ⇔ 1 + 4x + 4 x = 2x + 1
x≤0
√
⇔ 2 x = −x ⇔
4x = x2
⇔ x = 0.✭t❤ä❛
✤✐➲✉ ❦✐➺♥✮
❱➟② ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❧➔✿ x = 0✱ x = 71 ✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤✿
(6x + 10)
√
3x + 4 − (10x + 14)
√
x + 2 = 1✳
●ñ✐ þ✳ ❚❛ ❣✐↔ sû✿
6x + 10 = a (3x + 4) + b (x + 2)
10x + 14 = c (3x + 4) + d (x + 2)
⇔
6x + 10 = (3a + b) x + (4a + 2b)
10x + 14 = (3c + d) x + (4c + 2d)
✣ç♥❣ ♥❤➜t ❤➺ sè t❛ ✤÷ñ❝✿
3a + b = 6
4a + 2b = 10
⇔
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
a=1
b=3
✈➔
✷✵
3c + d = 10
4c + 2d = 14
⇔
c=3
d=1
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✳ ❑❤✐ ✤â ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t÷ì♥❣ ✤÷ì♥❣
▲í✐ ❣✐↔✐✳ ✣✐➲✉ ❦✐➺♥✿ x ≥ −4
3
✈î✐✿
√
√
[(3x + 4) + 3 (x + 2)] 3x + 4 − [3 (3x + 4) + (x + 2)] x + 2 = 1 ✭✶✮
√
√
✣➦t u = 3x + 4 ✈➔ v = x + 2✳ ❚❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝✿
u2 + 3v 2 u − 3u2 + v 2 v = 1 ⇔ u3 − 3u2 v + 3uv 2 − v 3 = 1
⇔ (u − v)3 = 1 ⇔ u − v = 1 ⇔ u = v + 1
❱➟②✿
√
x + 2 + 1 ⇔ 3x + 4 = x + 3 + 2 x + 2
x ≥ −1
√
2
⇔ 2x + 1 = 2 x + 2 ⇔
4x2 + 4x + 1 = 4 (x + 2)
√
x ≥ −1
7
2 ⇔x=
.✭t❤ä❛ ♠➣♥ ✤✐➲✉ ❦✐➺♥✮
⇔
4x2 = 7
2
√
± 7
◆❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❧➔✿ x = 2 ✳
√
3x + 4 =
√
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤✿ 2x2 − 11x + 21 − 3√4x − 4 = 0✳
3
●ñ✐ þ✳ ❉ò♥❣ ❤➺ sè ❜➜t ✤à♥❤ t❛ t➻♠ α✱ β ✱ γ s❛♦ ❝❤♦
2x2 − 11x + 21 = α(4x − 4)2 + β (4x − 4) + γ
⇔ 2x2 − 11x + 21 = 16αx2 + (4β − 32α) x + (16α − 4β + γ)
16α = 2
1 −7
⇔
⇔ (α; β; γ) =
;
; 12
4β − 32α = −11
8
4
16α − 4β + γ = 21
❚❛ ❝â ❧í✐ ❣✐↔✐✳
❚➟♣ ①→❝ ✤à♥❤✿ D = R✳ P❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t÷ì♥❣ ✤÷ì♥❣ ✈î✐
√
1
7
(4x − 4)2 − (4x − 4) + 12 − 3 4x − 4 = 0
✭✶✮
8
4
3
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✷✶
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✣➦t t =
√
3
4x − 4✱
t❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝ t6 − 14t3 − 24t2 + 96 = 0✱ ❤❛②
(t − 2)2 t4 + 4t3 + 12t2 + 18t + 24 = 0
✭✷✮
◆➳✉ t ≤ 0 t❤➻ t6 − 14t3 − 24t2 + 96 > 0✳ ◆➳✉ t > 0 t❤➻
t4 + 4t3 + 12t2 + 18t + 24 > 0✳
❉♦ ✤â (2) ⇔ t = 2 ⇒ x = 3.
◆❤÷ ✈➟②✱ sû ❞ö♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➺ sè ❜➜t ✤à♥❤ ❝❤♦ t❛ ❧í✐ ❣✐↔✐ ❜➔✐ t♦→♥
♠ët ❝→❝❤ r➜t tü ♥❤✐➯♥ ✈➔ rã r➔♥❣✳
✶✳✹ P❤÷ì♥❣ ♣❤→♣ ✤÷❛ ✈➲ ❤➺
✣➸ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ ❜➡♥❣ ❝→❝❤ ✤÷❛ ✈➲ ❤➺ ♣❤÷ì♥❣ tr➻♥❤ t❛ t❤÷í♥❣
✤➦t ➞♥ ♣❤ö✱ ♣❤➨♣ ✤➦t ➞♥ ♣❤ö ♥➔② ❝ò♥❣ ✈î✐ ♣❤÷ì♥❣ tr➻♥❤ tr♦♥❣ ❣✐↔ t❤✐➳t
❝❤♦ t❛ ♠ët ❤➺ ♣❤÷ì♥❣ tr➻♥❤✳ ❙❛✉ ✤➙②✱ t❛ s➩ tr➻♥❤ ❜➔② ♣❤÷ì♥❣ ♣❤→♣
❣✐↔✐ ♥➔② t❤æ♥❣ q✉❛ ❝→❝ ❜➔✐ t♦→♥ ✈î✐ ♥❤ú♥❣ ❝❤ó þ ✈➔ ❧í✐ ❣✐↔✐ ♥❣❛② s❛✉
✤â✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤✿
√
3+x+
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥✿
✣➦t✿
u=
v=
√
√
3+x
6−x
√
6−x−
3+x≥0
6−x≥0
(3 + x) (6 − x) = 3
⇔ −3 ≤ x ≤ 6.
✭✶✮
✱ u, v ≥ 0✱ s✉② r❛ u2 + v2 = 9.
❑❤✐ ✤â ♣❤÷ì♥❣ tr➻♥❤ ✤÷ñ❝ ❝❤✉②➸♥ t❤➔♥❤ ❤➺ ✿
u2 + v 2 = 9
u + v − uv = 3
⇔
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
(u + v)2 − 2uv = 9
u + v = 3 + uv
✷✷
⇒ (3 + uv)2 − 2uv = 9
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
⇔
⇔
uv = 0
uv = −4✭❧♦↕✐✮
√
3+x=0
⇔
√
6−x=0
u=0
⇔
v=0
x = −3
x = 6.
✭t❤ä❛ ♠➣♥ ✭✶✮✮
❱➟② ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❧➔ x = −3✱ x = 6✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
●ñ✐ þ✳ ✣è✐ ✈î✐ ♣❤÷ì♥❣ tr➻♥❤
❣✐↔✐✿
✣➦t u =
n
a − f (x)✱ v =
m
n
√
4
18 + 5x +
a + f (x) +
m
√
4
64 − 5x = 4✳
b − f (x) = c✱ t❛ ❝â ❝→❝❤
u+v =c
b + f (x)✱ ❞➝♥ ✤➳♥ ❤➺
un + v m = a + b
▲í✐ ❣✐↔✐✳ ❱î✐ ✤✐➲✉ ❦✐➺♥
x ≥ − 18
18 + 5x ≥ 0
5
⇔
64
x≤
64 − 5x ≥ 0
5
18
64
⇔ − ≤ x ≤ (∗)
5
5
√
√
u = 4 18 + 5x, v = 4 64 − 5x✱ ✈î✐ u ≥ 0, v ≥ 0✳
u4 = 18 + 5x
✣➦t
❙✉② r❛
v 4 = 64 − 5x
P❤÷ì♥❣ tr➻♥❤
✤➣ ❝❤♦ t÷ì♥❣ ✤÷ì♥❣
✈î✐ ❤➺✿
u+v =4
u+v =4
2
u4 + v 4 = 82 ⇔
u2 + v 2 − 2(uv)2 = 82
v ≥ 0, v ≥ 0
v ≥ 0, v ≥ 0
✣➦t ❙ ❂ ✉ ✰ ✈ ✈➔ P ❂ ✉✳✈✱ t❛ ❝â✿
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✷✸
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
S=4
S 2 − 2P
⇒
− 2P 2 = 82
P ≥ 0, S ≥ 0
S=4
P 2 − 32P + 87 = 0 ⇔
•
2
P ≥0
S=4
P = 3 ∨ P = 29
P ≥0
❱î✐ ❙ ❂ ✹✱ P ❂ ✸✱ ✉ ✈➔ ✈ ❧➔ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤✿
y=1
y 2 − 4y + 3 = 0 ⇔
❉♦ ✤â t❛ ❝â✿
❙✉② r❛
u=1
v=3
√
4
√
4
u=3
∨
v=1
18 + 5x = 1
64 − 5x = 3
⇔
y=3
√
4
∨
18 + 5x = 1
64 − 5x = 81
√
4
18 + 5x = 3
64 − 5x = 1
18 + 5x = 81
∨
64 − 5x = 1
✭t❤ä❛ (∗)✮
• ❱î✐ ❙ ❂ ✹✱ P ❂ ✷✾✱ ❦❤æ♥❣ tç♥ t↕✐ ✉ ✈➔ ✈✳
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ✷ ♥❣❤✐➺♠ ❧➔ x1 = − 175 , x2 = 635 ✳
⇔x=−
63
17
∨x=
5
5
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ √3x + 1 = −4x2 + 13x − 5.
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ 3x + 1 ≥ 0 ⇔ x ≥ − 31 . ✭✯✮
❱✐➳t ❧↕✐ ♣❤÷ì♥❣ tr➻♥❤ ❞÷î✐ ❞↕♥❣✿
√
3x + 1 = −(2x − 3)2 + x + 4 ⇔
✣➦t −2y + 3 =
√
3x + 1✱
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
√
3x + 1 = −(−2x + 3)2 + x + 4
✤✐➲✉ ❦✐➺♥ −2y + 3 ≥ 0 ⇔ y ≤ 23 .
✷✹
✭✯✯✮
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❑❤✐ ✤â ♣❤÷ì♥❣ tr➻♥❤ ✤÷ñ❝ ❝❤✉②➸♥ t❤➔♥❤ ❤➺ ✿
−2y + 3 = −(−2x + 3)2 + x + 4
(−2y + 3)2 = 3x + 1
⇔
(−2x + 3)2 = x + 2y + 1 (1)
(−2y + 3)2 = 3x + 1
(2)
⇒ (2x − 3)2 − (2y − 3)2 = −2 (x − y)
⇔ 4 (x − y) (x + y − 3) = −2 (x − y)
⇔ (x − y) (2x + 2y − 5) = 0 ⇔
•
2y = 5 − 2x
❱î✐ x = y t❤❛② ✈➔♦ ✭✶✮ ✤÷ñ❝✿
√
15 − 97
4x − 15x + 8 = 0 ⇔ x =
8
✭❞♦ ✤✐➲✉ ❦✐➺♥ ✭✯✮ ✈➔ ✭✯✯✮✮
√
11
−
73
4x2 − 11x + 3 = 0 ⇔ x =
8
✭❞♦ ✤✐➲✉ ❦✐➺♥ ✭✯✮ ✈➔ ✭✯✯✮✮
2
•
x=y
❱î✐ 2y = 5 − 2x t❤❛② ✈➔♦ ✭✶✮ ✤÷ñ❝✿
❱➟②
√
√
15 − 97
11 − 73
♣❤÷ì♥❣ tr➻♥❤ ❝â ❤❛✐ ♥❣❤✐➺♠ x = 8 ✱ x = 8 ✳
√
❚↕✐ s❛♦ t❛ ❧↕✐ ❝â ♣❤➨♣ ✤➦t −2y + 3 = 3x + 1 ❄
▲÷✉ þ✳
❚❤➟t ✈➟②✱ ✤è✐ ✈î✐ ♣❤÷ì♥❣ tr➻♥❤ ❝❤ù❛ ❝➠♥ t❤ù❝✱ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣
✤÷❛ ✈➲ ❤➺ ♥❤÷ tr➯♥ t❛ ❝â ✷ ❤÷î♥❣ ❝ì ❜↔♥ s❛✉✿
❉↕♥❣ ✶✳ P❤÷ì♥❣ tr➻♥❤ ❝❤ù❛ ❝➠♥ ❜➟❝ ❤❛✐ ✈➔ ❧ô② t❤ø❛ ❜➟❝ ❤❛✐✳
√
ax + b = c(dx + e)2 + αx + β
✈î✐ d = ac + α ✈➔ e = bc + β (∗)✳
P❤÷ì♥❣ ♣❤→♣ ❣✐↔✐✿
✣✐➲✉ ❦✐➺♥ ax + b ≥ 0✳
√
✣➦t✿ dy + e = ax + b✱ ✤✐➲✉ ❦✐➺♥ dy + e ≥ 0✳
❑❤✐ ✤â ♣❤÷ì♥❣ tr➻♥❤ ✤÷ñ❝ ❝❤✉②➸♥ t❤➔♥❤ ❤➺✿
dy + e =
√
ax + b
2
dy + e = c(dx + e) + αx + β
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
⇔
✷✺
(dy + e)2 = ax + b
c(dx + e)2 = −αx + dy + e − β
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
⇔
c(dy + e)2 = acx + bc
(1)
c(dx + e)2 = (ac − d) x + dy + bc (2)
▲➜② ✭✷✮ trø ✭✶✮ t❤❡♦ ✈➳✱ t❛ ✤÷ñ❝✿
d (x − y) .h (x, y) = 0 ⇔
x=y
(3)
h (x, y)
(4)
❱î✐ ✭✸✮ t❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝ ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❤❛✐ t❤❡♦ x✳
• ❱î✐ ✭✹✮ t❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝ ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❤❛✐ t❤❡♦ x✳
◆❤➟♥ ①➨t✳ ✣➸ sû ❞ö♥❣ ✤÷ñ❝ ♣❤÷ì♥❣ ♣❤→♣ tr➯♥ ❝➛♥ ♣❤↔✐ ❦❤➨♦ ❧➨♦ ❜✐➳♥
✤ê✐ ♣❤÷ì♥❣ tr➻♥❤ ❜❛♥ ✤➛✉ ✈➲ ❞↕♥❣ t❤ä❛ ♠➣♥ ✤✐➲✉ ❦✐➺♥ (∗)✳
❉↕♥❣ ✷✳ P❤÷ì♥❣ tr➻♥❤ ❝❤ù❛ ❝➠♥ ❜➟❝ ❜❛ ✈➔ ❧ô② t❤ø❛ ❜➟❝ ❜❛✳
•
√
3
ax + b = c(dx + e)3 + αx + β
✈î✐ d = ac + α ✈➔ e = bc + β ✳
P❤÷ì♥❣ ♣❤→♣ ❣✐↔✐✿
dy + e =
√
3
ax + b
❑❤✐ ✤â ♣❤÷ì♥❣ tr➻♥❤ ✤÷ñ❝ ❝❤✉②➸♥ t❤➔♥❤ ❤➺✿
dy + e =
√
3
ax + b
dy + e = c(dx + e)3 + αx + β
⇔
⇔
(dy + e)3 = ax + b
c(dx + e)3 = −αx + dy + e − β
c(dy + e)3 = acx + bc
(1)
c(dx + e)3 = (ac − d) x + dy + bc (2)
▲➜② ✭✷✮ trø ✭✶✮ t❤❡♦ ✈➳✱ t❛ ✤÷ñ❝✿
d (x − y) .h (x, y) = 0 ⇔
x=y
(3)
h (x, y)
(4)
❱î✐ ✭✸✮ t❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝ ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ t❤❡♦ x✳
• ❱î✐ ✭✹✮ t❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝ ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ t❤❡♦ x✳
•
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✷✻
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✶✳✺ P❤÷ì♥❣ ♣❤→♣ ❤➔♠ sè
❙û ❞ö♥❣ ❝→❝ t➼♥❤ ❝❤➜t ❝õ❛ ❤➔♠ sè ✤➸ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ ❧➔
♣❤÷ì♥❣ ♣❤→♣ ❦❤→ q✉❡♥ t❤✉ë❝✳ ❚❛ ❝â ❜❛ ❤÷î♥❣ →♣ ❞ö♥❣ s❛✉✿
❍÷î♥❣ ✶✿ ❚❤ü❝ ❤✐➺♥ t❤❡♦ ❝→❝ ❜÷î❝✿
❇÷î❝ ✶✳ ❈❤✉②➸♥ ♣❤÷ì♥❣ tr➻♥❤ ✈➲ ❞↕♥❣✿ f (x) = k
❇÷î❝ ✷✳ ❳➨t ❤➔♠ sè y = f (x)
❉ò♥❣ ❧➟♣ ❧✉➟♥ ❦❤➥♥❣ ✤à♥❤ ❤➔♠ sè ❧➔ ✤ì♥ ✤✐➺✉ ✭❣✐↔ sû ❧➔ ✤ç♥❣ ❜✐➳♥✮✳
❇÷î❝ ✸✳◆❤➟♥ ①➨t✿
✲ ❱î✐ x = x0 ⇔ f (x) = f (x0) = k✱ ❞♦ ✤â x = x0 ❧➔ ♥❣❤✐➺♠✳
✲ ❱î✐ x > x0 ⇔ f (x) > f (x0) = k✱ ❞♦ ✤â ♣❤÷ì♥❣ tr➻♥❤ ✈æ
♥❣❤✐➺♠✳
✲ ❱î✐ x < x0 ⇔ f (x) < f (x0) = k✱ ❞♦ ✤â ♣❤÷ì♥❣ tr➻♥❤ ✈æ
♥❣❤✐➺♠✳
❱➟② x0 ❧➔ ♥❣❤✐➺♠ ❞✉② ♥❤➜t ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤✳
❍÷î♥❣ ✷✿ ❚❤ü❝ ❤✐➺♥ t❤❡♦ ❝→❝ ❜÷î❝✿
❇÷î❝ ✶✳ ❈❤✉②➸♥ ♣❤÷ì♥❣ tr➻♥❤ ✈➲ ❞↕♥❣✿ f (x) = g (x)
❇÷î❝ ✷✳ ❳➨t ❤➔♠ sè y = f (x) ✈➔ y = g (x)
❉ò♥❣ ❧➟♣ ❧✉➟♥ ❦❤➥♥❣ ✤à♥❤ ❤➔♠ sè y = f (x) ✤ç♥❣ ❜✐➳♥✱ ❝á♥ ❤➔♠ sè
y = g (x) ❧➔ ❤➔♠ ❤➡♥❣ ❤♦➦❝ ♥❣❤à❝❤ ❜✐➳♥✳
❳→❝ ✤à♥❤ x0 s❛♦ ❝❤♦ f (x0) = g (x0)✳
❇÷î❝ ✸✳ ❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ ❞✉② ♥❤➜t x = x0✳
❍÷î♥❣ ✸✿ ❚❤ü❝ ❤✐➺♥ t❤❡♦ ❝→❝ ❜÷î❝✿
❇÷î❝ ✶✳ ❈❤✉②➸♥ ♣❤÷ì♥❣ tr➻♥❤ ✈➲ ❞↕♥❣✿ f (u) = f (v)
❇÷î❝ ✷✳ ❳➨t ❤➔♠ sè y = f (x)✳ ❉ò♥❣ ❧➟♣ ❧✉➟♥ ❦❤➥♥❣ ✤à♥❤ ❤➔♠
sè ❧➔ ✤ì♥ ✤✐➺✉ ✭❣✐↔ sû ❧➔ ✤ç♥❣ ❜✐➳♥✮
❇÷î❝ ✸✳ ❑❤✐ ✤â f (u) = f (v) ⇔ u = v, ∀u, v ∈ Df
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ √3x + 1 + x + √7x + 2 = 4.
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✷✼
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
●ñ✐ þ✳ ❱î✐ ❜➔✐ t♦→♥ ♥➔②✱ ♥➳✉ ❣✐↔✐ t❤❡♦ ❝→❝❤ t❤æ♥❣ t❤÷í♥❣ ♥❤÷ ❜➻♥❤
♣❤÷ì♥❣ ❤❛② ✤➦t ➞♥ ♣❤ö t❤➻ s➩ ❣➦♣ ♥❤✐➲✉ ❦❤â ❦❤➠♥✳ ❚✉② ♥❤✐➯♥✱ ♥➳✉
t✐♥❤ þ ♠ët ❝❤ót ❝❤ó♥❣ t❛ s➩ t❤➜② ♥❣❛② ✈➲ tr→✐ ❧➔ ♠ët ❤➔♠ ✤ç♥❣ ❜✐➳♥
✈➔ x = 1 ❧➔ ♠ët ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤✳ ◆➯♥ t❤❡♦ ✤à♥❤ ❧➼ ✶✱ t❛ ❝â
✤÷ñ❝ x = 1 ❧➔ ♥❣❤✐➺♠ ❞✉② ♥❤➜t✳ ❱➟② t❛ ❝â ❧í✐ ❣✐↔✐ ♥❤÷ s❛✉✿
▲í✐ ❣✐↔✐✳ ✣✐➲✉ ❦✐➺♥ D =
x ∈ R|x ≥
√
❳➨t ❤➔♠ sè f (x) = 3x + 1 + x +
❚❛ ❝â f ❧➔ ❤➔♠ ❧✐➯♥ tö❝ tr➯♥ ❉ ✈➔
7−
√
√
57
2
✳
7x + 2✳
7
1+ √
3
2 7x + 2
+
f (x) = √
>0
√
2 3x + 1 2 x + 7x + 2
♥➯♥ ❤➔♠ sè f (x) ❧✉æ♥ ✤ç♥❣
❜✐➳♥✳
▼➦t ❦❤→❝✱ t❛ t❤➜② f (1) = 4✳
∗ ◆➳✉ x > 1 s✉② r❛ f (x) > f (1) = 4 ♥➯♥ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ✈æ
♥❣❤✐➺♠✳
∗ ◆➳✉ x < 1 s✉② r❛ f (x) < f (1) = 4 ♥➯♥ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ✈æ
♥❣❤✐➺♠✳
❱➟② x = 1 ❧➔ ♥❣❤✐➺♠ ❞✉② ♥❤➜t ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐
♣❤÷ì♥❣ tr➻♥❤
√
√
3
x+2+
3
x+1=
√
3
2x2 + 1 +
√
3
2x2 ✳
●ñ✐ þ✳ ❱î✐ ✤÷í♥❣ ❧è✐ ♥❤÷ ❜➔✐ t♦→♥ tr➯♥ t❤➻ t❛ s➩ ❦❤â ❦❤➠♥ ✤➸ ❣✐↔✐ ❜➔✐
t♦→♥ ♥➔②✳ ❚✉② ♥❤✐➯♥✱ t❛ ✤➸ þ t❤➜② ❜✐➸✉ t❤ù❝ ❞÷î✐ ❞➜✉ ❝➠♥ ð ❝↔ ❤❛✐ ✈➳
❝â ✤➦❝ ✤✐➸♠ ❧➔ x + 2 = (x + 1) + 1 ✈➔ 2x2 + 1 = (2x2) + 1✳ ❉♦ ✈➟②✱ ♥➳✉
√
√
t❛ ✤➦t u = x + 1, v = 2x2 t❤➻ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ trð t❤➔♥❤✿
√
√
u3 + 1 + u = v 3 + 1 + v ⇔ f (u) = f (v)✳
√
tr♦♥❣ ✤â t❛ ❞➵ t❤➜② f (t) = t3 + 1 + t ❧➔ ♠ët ❤➔♠ ❧✐➯♥ tö❝ ✈➔ ✤ç♥❣
3
3
3
3
3
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✷✽
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❜✐➳♥ tr➯♥ t➟♣ ①→❝ ✤à♥❤ ❝õ❛ ♥â✳ ◆➯♥ t❤❡♦ ✤à♥❤ ❧➼ ✶ t❛ ❞➝♥ ✤➳♥ ♠ët
♣❤÷ì♥❣ tr➻♥❤ ✤ì♥ ❣✐↔♥ ❤ì♥ ❧➔ u = v✳
▲í✐ ❣✐↔✐✳ ❚➟♣ ①→❝ ✤à♥❤ D
= R✳
✣➦t
♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ trð t❤➔♥❤✿
√
3
t
3
❉♦ ✤â✱
(t3 + 1)
x + 1, v =
√
3
2x2
t❤➻
√
3
tö❝ ✈➔ ❝â
+ 1 > 0 ✈î✐ ♠å✐ t✱ ♥➯♥ f (t) ❧✉æ♥ ✤ç♥❣ ❜✐➳♥✳
2
tr♦♥❣ ✤â t❛ ❞➵ t❤➜②
f (t) =
√
3
v 3 + 1 + v ⇔ f (u) = f (v)✳
√
f (t) = 3 t3 + 1 + t ❧➔ ♠ët ❤➔♠ ❧✐➯♥
u3 + 1 + u =
2
u =
2
f (u) = f (v) ⇔ u = v ⇔ 2x = x + 1 ⇔
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ✷ ♥❣❤✐➺♠ ❧➔ x = 1 ✈➔
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
32x
3
−x+2
− 3x
3
+2x
x=1
−1
x=
.
2
−1
x=
✳
2
+ x3 − 3x + 2 = 0.
●ñ✐ þ✳ ❚❛ ❝➛♥ t➻♠ k s❛♦ ❝❤♦
−(x3 − 3x + 2) = k
x3 + 2x − 2x3 − x + 2
⇒ k = 1.
❈á♥ ✤è✐ ✈î✐ ♣❤÷ì♥❣ tr➻♥❤ tê♥❣ q✉→t af (x) − ag(x) = h (x) ✤÷ñ❝ ❣✐↔✐
t÷ì♥❣ tü✳ ❚❤÷í♥❣ t❤➻ t❛ ❞ò♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➺ sè ❜➜t ✤à♥❤ ♥❤÷ tr➯♥
✤➸ ✤÷❛ ✈➲
h (x) = k [g (x) − f (x)] .
▲í✐ ❣✐↔✐✳ P❤÷ì♥❣ tr➻♥❤ ✭✶✮ ✈✐➳t ❧↕✐
3
32x
⇔ 32x
−x+2
3
−x+2
− 3x
3
+2x
= − 2x3 − x + 2 + x3 + 2x
+ 2x3 − x + 2 = 3x
3
+2x
✭✷✮
+ x3 + 2x
⇔ f 2x3 − x + 2 = f x3 + 2x
✱ ✈î✐ f (t) = 3t + t
✭✸✮
❍➔♠ sè f ✤ç♥❣ ❜✐➳♥ tr➯♥ ❘ ✈➻ f (t) = 3t ln 3 + 1 > 0, ∀t ∈ R✳ ❱➟②
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✷✾
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
(3) ⇔ 2x3 − x + 2 = x3 + 2x ⇔ x3 − 3x + 2 = 0 ⇔
x = −2
x = 1.
P❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ❤❛✐ ♥❣❤✐➺♠ x = −2, x = 1.
❇➔✐ t♦→♥ ✹✳ ✭❍❙● ❚❤→✐ ❇➻♥❤ ♥➠♠ ❤å❝ ✷✵✶✵✲✷✵✶✶✮✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤✿
log3
2x − 1
2
2 = 3x − 8x + 5.
(x − 1)
✭✶✮
●ñ✐ þ✳ ❚❛ ❝➛♥ t➻♠ α, β, γ s❛♦ ❝❤♦
3x2 −8x + 5 = α (2x − 1) +
β(x − 1)2 + 1
α = −1
β=3
⇒
β=3
2α − 2β = −8 ⇔
γ = 1.
−α + β + γ = 5
(x)
❉↕♥❣ tê♥❣ q✉→t loga fg(x)
= h(x) ✤÷ñ❝ ❣✐↔✐ t÷ì♥❣ tü✳ ❚❛ t❤÷í♥❣
❞ò♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➺ sè ❜➜t ✤à♥❤ ♥❤÷ tr➯♥ ✤➸ ✤÷❛ ✈➲ ♠ët tr♦♥❣
❝→❝ tr÷í♥❣ ❤ñ♣ h(x) = −f (x) + ak g (x) + k✱ h (x) = f (x) − ak g (x) + k✱
h (x) = k [g (x) − f (x)]✳
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ 0, 5 < x = 1. ❑❤✐ ✤â ✭✶✮ t÷ì♥❣ ✤÷ì♥❣
log3 (2x − 1) − log3 (x − 1)2 = −(2x − 1) + 3(x − 1)2 + 1
✭✷✮
⇔ log3 (2x − 1) + (2x − 1) = log3 (3(x − 1)2 ) + 3(x − 1)2
⇔ f (2x − 1) = f 3(x − 1)2 ,
❱➻ f
❝â
(t) =
1
+ 1 > 0, ∀t > 0
ln 3
✈î✐ f (t) = log3t + t. ✭✸✮
♥➯♥ f ✤ç♥❣ ❜✐➳♥ tr➯♥ (0; +∞)✱ tø ✭✸✮
2x − 1 = 3(x − 1)2 ⇔ 3x2 − 8x + 4 ⇔ x ∈
❚➟♣ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✭✶✮ ❧➔ S =
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✸✵
2
3
2
2,
3
✭t❤ä❛ ✤✐➲✉ ❦✐➺♥✮✳
2,
✳
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✶✳✻ P❤÷ì♥❣ ♣❤→♣ ❤➡♥❣ sè ❜✐➳♥ t❤✐➯♥
❳➨t ♣❤÷ì♥❣ tr➻♥❤ ➞♥ x✱ t❤❛♠ sè m✿ f (x; m) = 0✳ ❚✉② ♥❤✐➯♥✱ tr♦♥❣
♣❤÷ì♥❣ ♣❤→♣ ♥➔② t❛ ❧↕✐ ❝♦✐ ➞♥ ❧➔ m✱ t❤❛♠ sè ❧➔ x✳ ●✐↔✐ m t❤❡♦ x rç✐
q✉❛② ❧↕✐ ➞♥ x✳ ❚❛ t❤÷í♥❣ ❞ò♥❣ ♣❤÷ì♥❣ ♣❤→♣ ♥➔② ❦❤✐ t❤❛♠ sè m ❝â
♠➦t ✈î✐ ❜➟❝ ❤❛✐ ✈➔ ❜✐➺t t❤ù❝ ∆ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❤❛✐ ➞♥ m ✤â ❧➔
sè ❝❤➼♥❤ ♣❤÷ì♥❣ ✭∆ ❧➔ ❜➻♥❤ ♣❤÷ì♥❣ ❝õ❛ ♠ët ❜✐➸✉ t❤ù❝✮✳ ❚r♦♥❣ ♠ët
sè tr÷í♥❣ ❤ñ♣✱ t❛ ❝á♥ ❝â t❤➸ ❝♦✐ sè ❧➔ ➞♥✳ ✣➙② ❧➔ ♠ët ♣❤÷ì♥❣ ♣❤→♣
r➜t ✤➦❝ ❜✐➺t✱ sü ❤✐➺♥ ❞✐➺♥ ❝õ❛ ♣❤÷ì♥❣ ♣❤→♣ ♥➔② ✤➣ ❣â♣ t❤➯♠ ♥❤ú♥❣
❧í✐ ❣✐↔✐ ✤ë❝ ✤→♦ ❝❤♦ ❝→❝ ❜➔✐ t♦→♥✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ (x + 1)2 + √x + 6 = 5
●✐↔✐✳ P❤÷ì♥❣ tr➻♥❤ ✈✐➳t ❧↕✐✿ (1) ⇔
(x + 1) + 5 = 5 − (x + 1)2
✣➦t y = x + 1✳ ❚❛ ❝â ✿
(2) ⇔
y + 5 = 5 − y2 ⇒ y + 5 = 5 − y2
2
✭✶✮
✭✷✮✳
✈î✐ ✤✐➲✉ ❦✐➺♥ 5 − y2 ≥ 0
⇒ y + 5 = 52 − 2y 2 .5 + y 4
✣➦t 5 = t t❛ ❝â✿
2
2
5 = t = y2 − y
4
t − 2y + 1 t + y − y = 0 ⇔
5 = t = y2 + y + 1
√
√
√
−
5
≤
y
≤
5
√
1
−
21
2
y
=
1
±
21
y −y−5=0
2 √
y=
⇔
⇒
2
2
√
−1 + 17
y +y−4=0
y=
−1
±
17
y=
2
2
❚ø ✤â s✉② r❛ ✿
√
21
−1 − 21
−1
x=
x=
2
2√
√
⇔
−1 + 17
−3 + 17
x=
−1
x=
2
2
1−
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
√
✸✶
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ✷ ❝â ♥❣❤✐➺♠✿
√
√
−1 − 21
−3 + 17
x=
✱x=
2
2
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
68 15
= .
✭✶✮
x3
x
√
◆➳✉ sû ❞ö♥❣ ❜✐➳♥ ✤ê✐ (1) ⇔ x6 − 15x2 + 2 17 = 0✳ ✣➦t
√
x2 = t > 0✱ t❛ ❝â ♣❤÷ì♥❣ tr➻♥❤ t3 − 15t + 2 17 = 0✳ ✣â ❧➔ ♠ët ♣❤÷ì♥❣
x3 +
▲÷✉ þ✳
tr➻♥❤ ❜➟❝ ❜❛✳ ❑❤æ♥❣ ❝â ♥❤➟♥ ①➨t ✈➲ ✤♦→♥ ♥❣❤✐➺♠ ✈æ t➾ ♥➯♥ ✈✐➺❝ ✤♦→♥
♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ♥➔② ✤➸ ❜✐➳♥ ✤ê✐ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ t➼❝❤ ✈➝♥
❝á♥ ❦❤â ❦❤➠♥✳ ❚❛ ♥❣❤➽ ✤➳♥ ✈✐➺❝ sû ❞ö♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➡♥❣ sè ❜✐➳♥
t❤✐➯♥✳ ▲í✐ ❣✐↔✐ ♥❤÷ s❛✉✿
▲í✐ ❣✐↔✐✳ ✣✐➲✉ ❦✐➺♥✿ x = 0✳
•x
❧➔ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✭✶✮ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐
√
❙✉② r❛
√
√
68
15
17 17 − 2
2
x3 + 3 =
=
⇔ x3 +
x
x
x3
x
17
❧➔ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ s❛✉ ✈î✐ ➞♥ ❧➔ a✿
2a a2 − 2
x + 3 =
⇔ x2 a2 − 2a − x6 − 2x2 = 0
x
x
a = −x2
(2)
⇔
2 + x4
a=
(3)
x2
√
√
a = 17 ✈➔♦ ✭✷✮✱ t❛ ✤÷ñ❝ 17 = −x2 ✳ P❤÷ì♥❣ tr➻♥❤
3
❚❤❛②
♥❣❤✐➺♠✳
√
• ❚❤❛② a = 17 ✈➔♦ ✭✸✮✱ t❛ ✤÷ñ❝
•
√
√
2 + x4
4
17 =
⇔
x
−
17x2 + 2 = 0 ⇔ x2 =
2
x
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ❜è♥ ♥❣❤✐➺♠ ❧➔ x1,2 =
√
x3,4 = −
√
√
♥➔② ✈æ
17 ± 3
.
2
17 ± 3
✱
2
17 ± 3
✳
2
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✸✷
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
x+
11 +
√
x = 11
✭✶✮
▲÷✉ þ✳ ❑❤✐ ❝â sü ❧➦♣ ❝õ❛ sè ♠ô ð ❧ô② t❤ø❛ ✈î✐ ❝→❝ ❝ì sè ❦❤→❝ ♥❤❛✉
❤♦➦❝ ❝â sü ❧➦♣ ❝õ❛ ❝→❝ ❤➡♥❣ sè ❞÷î✐ ❝→❝ ❝➠♥ t❤ù❝ ❝â ❜➟❝ ❦❤→❝ ♥❤❛✉
t❤➻ ❜↕♥ ❝â t❤➸ ♥❣❤➽ ✤➳♥ ♣❤÷ì♥❣ ♣❤→♣ ❤➡♥❣ sè ❜✐➳♥ t❤✐➯♥✳ ❚❤❡♦ ❞ã✐ ❧í✐
❣✐↔✐ s❛✉ ✤➙②✿
▲í✐ ❣✐↔✐✳ ✣✐➲✉ ❦✐➺♥ 0 < x < 11✳
❱î✐ ✤✐➲✉ ❦✐➺♥ ✭✯✮ t❛ ❝â
(1) ⇔ 11 − x =
11 +
√
✭✯✮
x ⇔ (11 − x)2 = 11 +
√
x
✣➦t 11 = a✱ ♣❤÷ì♥❣ tr➻♥❤ ✭✷✮ ✤÷ñ❝ ✈✐➳t t❤➔♥❤
(a − x)2 = a +
√
x ⇔ a2 − 2ax + x2 = a +
√
⇔ a2 − (2x + 1) a + (x2 − x) = 0
√
✭✷✮
x
❳❡♠ ✭✸✮ ❧➔ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❤❛✐ ✤è✐ ✈î✐ a✳ ❚❛ ❝â✿
✭✸✮
√
√
√
2
∆ = (2x + 1)2 − 4(x2 − x) = 4x + 4 x + 1 = (2 x + 1)
√
√
⇒ ∆=2 x+1
❙✉② r❛ a1 = x + √x√+ 1✱ a2 = x − √x✳ ❉♦ ✤â✱
(4)
a=x+ x+1
(3) ⇔
√
a=x− x
(5)
❚❤❛② a = 11 t❛ ❝â✿
√
√
(4) ⇔ x√+ x + 1 = 11 ⇔ x + x − 10 = 0
√
41 − 1 √
⇔ x=
✭ x > 0✮
2 √
❙✉② r❛✿ x = 21 −2 41 ✭t❤ä❛ ♠➣♥ ✤✐➲✉ ❦✐➺♥✮
√
√
√
√
3 5+1 √
(5) ⇔ x − x = 11 ⇔ x − x − 11 = 0 ⇔ x =
✭ x > 0✮
2
√
❙✉② r❛✿ x = 23 +23 5 ✭t❤ä❛ ♠➣♥ ✤✐➲✉ ❦✐➺♥✮
√
√
21 − 41 23 + 3 5
❱➟② t➟♣ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❧➔ S =
;
2
2
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✸✸
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
✳
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
✶✳✼ P❤÷ì♥❣ ♣❤→♣ sû ❞ö♥❣ ✤à♥❤ ❧þ ▲❛❣r❛♥❣❡
✣➙② ❧➔ ♠ët ♣❤÷ì♥❣ ♣❤→♣ ✤➦❝ ❜✐➺t ♠î✐ ①✉➜t ❤✐➺♥ t❤í✐ ❣✐❛♥ ❣➛♥ ✤➙②✳
❈ì sð ❝õ❛ ♣❤÷ì♥❣ ♣❤→♣ ♥➔② ❧➔ ✤à♥❤ ❧➼ s❛✉✿
✣à♥❤ ❧þ✳ ✭✣à♥❤ ❧➼ ▲❛❣r❛♥❣❡✮✿ ◆➳✉ ❤➔♠ sè y = f (x) ❧✐➯♥ tö❝ tr➯♥ ✤♦↕♥
[a; b] ✈➔ ❝â ✤↕♦ ❤➔♠ tr➯♥ ❦❤♦↔♥❣ (a; b) t❤➻ tç♥ t↕✐ ♠ët sè c s❛♦ ❝❤♦
f (b) − f (a) = f (c) (b − a)✳
❙❛✉ ✤➙② t❛ s➩ tr➻♥❤ ❜➔② ♠ët ✈➔✐ ❞↕♥❣ ♣❤÷ì♥❣ tr➻♥❤ ✤÷ñ❝ ❣✐↔✐ ❜➡♥❣
❝→❝❤ ✈➟♥ ❞ö♥❣ ✤à♥❤ ❧þ ▲❛❣r❛♥❣❡✳ ❚✉② ♥❤✐➯♥✱ ❝á♥ ♥❤✐➲✉ ❞↕♥❣ ❦❤→❝ ♥ú❛
❝❤÷❛ ✤÷ñ❝ ✤➲ ❝➟♣ ✤➳♥✱ ✤➸ ❣✐↔✐ ❝❤ó♥❣ t❛ ❝➛♥ ♥❤➟♥ r❛ ✤÷ñ❝ ✤➦❝ t❤ò ❝õ❛
✤à♥❤ ❧þ ▲❛❣r❛♥❣❡ ✤÷ñ❝ ❝❤❡ ❣✐➜✉ tr♦♥❣ ♣❤÷ì♥❣ tr➻♥❤✳
❉↕♥❣ ✶✳ P❤÷ì♥❣ tr➻♥❤ ah(x) − bh(x) = k (a − b) h (x)✱ ✈î✐
0 < a = 1✱ 0 < b = 1✱a > b✱ k ≤ 0 ❤♦➦❝ k = 1✱ h(x) ①→❝ ✤à♥❤ tr➯♥
✤♦↕♥ [b; a]✳
P❤÷ì♥❣ ♣❤→♣✳ ❱✐➳t ❧↕✐
✭✯✮
❳➨t ❤➔♠ sè ✭❜✐➳♥ t✮✿ f (t) = th(x) − k.h (x) t✳ ❑❤✐ ✤â✱ tø ✭✯✮ t❛ ❝â
ah(x) − kah(x) = bh(x) − kbh(x)
f (a) = f (b) ⇔ f (a) − f (b) = 0.
❚❤❡♦ ✤à♥❤ ❧þ ▲❛❣r❛♥❣❡✱ tç♥ t↕✐ c ∈ (b; a) s❛♦ ❝❤♦
f (b) − f (a) = f (c) (b − a) ⇒ f (c) = 0.
❚ø ✤➙② t➻♠ ✤÷ñ❝ x s❛✉ ✤â t❤û ❧↕✐ ✤➸ ❝❤å♥ ♥❣❤✐➺♠✳
❇➔✐ t♦→♥ ✶✳ ✭✣➲ ♥❣❤à ❖❧②♠♣✐❝ ✸✵✴✵✹✴✷✵✶✵✮ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
✭✶✮
●✐↔✐✳ ❉➵ t❤➜② x = 0 ❧➔ ♠ët ♥❣❤✐➺♠ ❝õ❛ ✭✶✮✳ ●✐↔ sû α ❧➔ ♠ët ♥❣❤✐➺♠
❜➜t ❦➻ ❝õ❛ ✭✶✮✳ ❑❤✐ ✤â✱ 3cos α − 2cos α = cos α ⇔ 3cos α − 3 cos α =
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✸✹
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
3cos x − 2cos x = cos x.
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
2cos x − 2 cos α✳
✭✷✮✳ ❳➨t ❤➔♠ sè f (t) = tcos α − t cos α✱ ✈î✐ t > 1✳ ❍➔♠
sè f ❧✐➯♥ tö❝ tr➯♥ (1; +∞) ✈➔ ❝â f (t) = cos αtcos α−1 − cos α✳ ❚ø ✭✷✮ t❛
❝â f (2) = f (3)✳ ❍➔♠ f ❧✐➯♥ tö❝ tr➯♥ ✤♦↕♥ [2; 3] ✈➔ ❝â ✤↕♦ ❤➔♠ tr➯♥
❦❤♦↔♥❣(2; 3)✱ ❞♦ ✤â tç♥ t↕✐ b ∈ (2; 3) s❛♦ ❝❤♦
cos α−1
f (3) − f (2) = f (b) (2 − 3) ⇔ f (b) = 0 ⇔ cos
− cos α = 0
α.b π
α = + kπ
cos α = 0
cos α = 0
2
⇔
⇔
⇔
cos α−1
b
=1
cos α = 1
α = k2π.
◆❣❤✐➺♠ ❝õ❛ ✭✶✮ ❧➔ x = π2 + kπ✱ x = k2π(k ∈ Z)✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
xlog3 7 = 2log3 x + log3 x5
✭✶✮
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x > 0✳ ●✐↔ sû ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ x = α✱ tù❝ ❧➔
αlog3 7 = 2log3 α + log3 α5
⇔ 7log3 α − 2log3 α = (7 − 2) log3 α
⇔ 7log3 α − 7log3 α = 2log3 α − 2log3 α
(2)
❳➨t ❤➔♠ sè f (t) = tlog3α − tlog3α✱ ✈î✐ t > 0✳
❑❤✐ ✤â✱ tø ✭✷✮ t❛ ❝â
f (7) = f (2) ⇔ f (7) − f (2) = 0✳
❍➔♠ sè f ❧✐➯♥ tö❝ tr➯♥ ✤♦↕♥ [2; 7] ✈➔ ❝â ✤↕♦ ❤➔♠
f (t) = (log3 α) tlog3 α−1 − log3 α = tlog3 α−1 − 1 log3 α.
❚❤❡♦ ✤à♥❤ ❧þ ▲❛❣r❛♥❣❡✱ tç♥ t↕✐ c ∈ (2; 7) s❛♦ ❝❤♦
f (7) − f (2) = f (c) (7 − 2) ⇒ f (c) = 0,
♥❣❤➽❛ ❧➔
clog3 α−1 − 1 log3 α = 0 ⇔
⇔
log3 α = 0
log3 α = 1
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
⇔
log3 α = 0
clog3 α−1 = 1
α=1
α=3
✸✺
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❚❤❛② x = 1✱ x = 3 ✈➔♦ ✭✶✮ t❤➜② t❤ä❛ ♠➣♥✳ ❱➟② t➟♣ ♥❣❤✐➺♠ ❝õ❛ ✭✶✮ ❧➔
{1, 3}✳
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
7cot x − 11cot x = 12 cot x.
❍÷î♥❣ ❞➝♥✳ ✣✐➲✉ ❦✐➺♥ x = kπ, k ∈ Z✳ ●✐↔ sû α ❧➔ ♠ët ♥❣❤✐➺♠ ❝õ❛
♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦✳ ❑❤✐ ✤â✱
7cot α − 11cot α = 3 (11 − 7) cot α
⇔ 7cot α + 3.7 cot α = 11cot α + 3.11 cot α.
❘ç✐ ①➨t ❤➔♠ sè f (t) = tcot α + 3t cot α✳
❉↕♥❣ ✷✳ P❤÷ì♥❣ tr➻♥❤ (a + d)h(x) − ah(x) = (b + d)h(x) − bh(x)✱
✈î✐ 0 < a = 1✱ 0 < b = 1✱ b > a✱ d > 0✱ h(x) ①→❝ ✤à♥❤ tr➯♥ [b; a]
P❤÷ì♥❣ ♣❤→♣✳
❳➨t ❤➔♠ sè ❜✐➳♥ t✿ f (t) = (t + d)h(x) − th(x)✳ ❚ø ♣❤÷ì♥❣ tr➻♥❤ ✤➣
❝❤♦ t❛ ❝â f (a) = f (b) ⇔ f (a) − f (b) = 0. ❚❤❡♦ ✤à♥❤ ❧➼ ▲❛❣r❛♥❣❡✱
tç♥ t↕✐ c ∈ (b; a) s❛♦ ❝❤♦ f (b) − f (a) = f (c) (b − a) ⇒ f (c) = 0✳ ❚ø
✤➙② t❛ t➻♠ ✤÷ñ❝ x✱ s❛✉ ✤â t❤û ❧↕✐ ✤➸ ❝❤å♥ ♥❣❤✐➺♠✳
❇➔✐ t♦→♥ ✹✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
1
1
−
=
2x 3x
5
14
x
4
21
−
x
.
▲í✐ ❣✐↔✐✳ ❚➟♣ ①→❝ ✤à♥❤ R✳ P❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ✈✐➳t ❧↕✐
1
2
⇔
x
x
5
1
−
=
14
3
x
5
1
5
+
−
14 7
14
x
x
x
4
−
21
4
1
=
+
21 7
x
x
4
21
−
✭✶✮
●✐↔ sû ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ x = α✳ ❚ø ✭✶✮ t❛ ❝â
1
5
+
14 7
α
❳➨t ❤➔♠ sè f (t) =
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
−
α
5
14
1
t+
7
4
1
+
21 7
=
α
− tα ✱
α
−
4
21
α
(2)
✈î✐ t > 0✳ ❚ø ✭✷✮ t❛ ❝â
✸✻
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
5
14
f
=f
4
21
5
14
⇒f
❘ã r➔♥❣✱ ❤➔♠ sè f ❧✐➯♥ tö❝ tr➯♥ ✤♦↕♥
α−1
1
f (t) = α t +
7
4 5
;
∀t ∈
.
21 14
− αtα−1 = α
−f
4
21
= 0.
4 5
;
✈➔
21 14
α−1
1
t+
− tα−1
7
✱
❚❤❡♦ ✤à♥❤ ❧þ ▲❛❣r❛♥❣❡✱ tç♥ t↕✐ c ∈
5
14
f
−f
4
21
4 5
;
s❛♦ ❝❤♦
21 14
4
5
−
= f (c)
⇔ f (c) = 0
14 21
⇔α
⇔
1
c+
7
α−1
α−1
−c
α=0
=0⇔
α−1
1
c+
7
= cα−1
α=0
α = 1.
❚❤❛② x = 0✱ x = 1 ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t❤➜② t❤ä❛ ♠➣♥✳ ❱➟②
♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❧➔ x = 0✱ x = 1✳
❇➔✐ t♦→♥ ✺✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ 2log x +2log x
5
3
5
2
= x+xlog5 7 ✳
✭✯✮
▲í✐ ❣✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x > 0✳ ●✐↔ sû α ❧➔ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✭✯✮✳
❑❤✐ ✤â
3
2
2log5 α + 2log5 α = α + αlog5 7
⇔ 8log5 α + 4log5 α = 5log5 α + 7log5 α
⇔ 8log5 α − 5log5 α = 7log5 α − 4log5 α ✳
✭✶✮
❳➨t ❤➔♠ sè f (t) = (t + 3)log α − tlog α✱ ✈î✐ t > 0✳ ❑❤✐ ✤â✱ tø ✭✶✮ t❛ ❝â
5
5
f (5) = f (4) ⇔ f (5) − f (4) = 0
❍➔♠ f ❧✐➯♥ tö❝ tr➯♥ ✤♦↕♥ [4; 5] ✈➔
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✸✼
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
f (t) = (t + 3)log5 α−1 − tlog5 α−1 log5 α✱ ∀t ∈ (4; 5)✳
❚❤❡♦ ✤à♥❤ ❧þ ▲❛❣r❛♥❣❡✱ tç♥ t↕✐ c ∈ (4; 5) s❛♦ ❝❤♦
f (5) − f (4) = f (c) (5 − 4) ⇔ f (c) = 0
⇔ (c + 3)log5 α−1 − clog5 α−1 log5 α = 0
⇔
log5 α = 0
(c + 3)log5 α−1 = clog5 α−1
⇔
α=1
α = 5.
❚❤❛② x = 1✱ x = 5 ✈➔♦ ✭✯✮ t❤➜② ✤ó♥❣✳ ❱➟② S = {1, 5} .
✶✳✽ P❤÷ì♥❣ ♣❤→♣ ❤➻♥❤ ❤å❝
❙û ❞ö♥❣ ❦➳t q✉↔ r➜t q✉❡♥ t❤✉ë❝✿ ◆➳✉ ✤÷í♥❣ t❤➥♥❣ ∆ ✤✐ q✉❛ ✤✐➸♠
−
M (x0 ; y0 ) ✈➔ ❝â ✈❡❝tì ❝❤➾ ♣❤÷ì♥❣ →
u = (a; b) t❤➻ ∆ ❝â ♣❤÷ì♥❣ tr➻♥❤
t❤❛♠ sè✿
x = x0 + at
y = y0 + bt
(t ∈ R)
❑➳t q✉↔ ✈æ ❝ò♥❣ ✤ì♥ ❣✐↔♥ tr➯♥ ❧↕✐ ❝❤♦ t❛ ♠ët ♣❤➨♣ ✤➦t ➞♥ ♣❤ö r➜t
✤➭♣ ❝ô♥❣ ♥❤÷ ♠ët ❝→❝❤ s→♥❣ t→❝ ✤➲ t♦→♥ r➜t ♥❤❛♥❤ ❝❤â♥❣ ✭s→♥❣ t→❝
✤➲ t♦→♥ s➩ ✤÷ñ❝ ✤➲ ❝➟♣ ð ❝❤÷ì♥❣ ✷✮✳
❱➼ ❞ö✳ √x3 + 8 + 3√12 − x3 = 10
✭✯✮
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥
x3 + 8 ≥ 0
12 − x3 ≥ 0
√
x3 + 8 = 1 + 3t✳
⇔ −2 ≤ x ≤
√
3
12✳
✣➦t
√
❚❤❛② ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t❛ ✤÷ñ❝ 12 − x3 = 3 − t ✈➔ ✤✐➲✉ ❦✐➺♥
❝õ❛ t ❧➔ −1
≤ t ≤ 3✳ ❱➟② t❛ ❝â
3
√
√
x3 + 8 = 1 + 3t
12 − x3 = 3 − t
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
⇔
x3 + 8 = (1 + 3t)2
(i)
12 − x3 = (3 − t)2
(ii)
✸✽
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❈ë♥❣ ✭✐✮ ✈➔ ✭✐✐✮ t❤❡♦ ✈➳ t❛ ✤÷ñ❝
20 = 1 + 6t + 9t2 + 9 − 6t + t2 ⇔ 10t2 = 10 ⇔
t=1
✭❧♦↕✐✮
√
❱➟② x3 + 8 = 4 ⇔ x3 + 8 = 16 ⇔ x3 = 8 ⇔ x = 2 ✭t❤ä❛ ♠➣♥ ✤✐➲✉
❦✐➺♥✮✳
▲÷✉ þ✳ ❇➼ q✉②➳t ♥➔♦ ❞➝♥ ✤➳♥ ♣❤➨♣ ✤➦t √x3 + 8 = 1 + 3t ❄ ✳✣â ❝❤➼♥❤
❧➔ tø ♣❤÷ì♥❣ tr➻♥❤ t❤❛♠ sè ❝õ❛ ✤÷í♥❣ t❤➥♥❣ ∆✳ ❳➨t ✤÷í♥❣ t❤➥♥❣ ∆
−
✤✐ q✉❛ ✤✐➸♠ A (1; 3) ✈➔ ❝â ✈➨❝✲tì ❝❤➾ ♣❤÷ì♥❣ →
u = (3; −1)✳ ❚❤➻ ♣❤÷ì♥❣
tr➻♥❤ tê♥❣ q✉→t ❝õ❛ ∆ ❧➔
√
√
(x − 1) + 3 (y − 3) = 0✳ ❈❤å♥ x = x3 + 8✱ y = 12 − x3 ✱ t❤➻ t❛ ✤÷ñ❝
♣❤÷ì♥❣ tr➻♥❤
x3 + 8 − 1 + 3
⇔
t = −1
12 − x3 − 3 = 0
x3 + 8 + 3 12 − x3 = 10
√
◆❤÷ ✈➟②✱ ✤➸ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ t❛ ✤➦t x3 + 8 = 1 + 3t✳ ✣è✐ ✈î✐ ❜➔✐
√
t♦→♥ tr➯♥ ❝á♥ ❝â ❝→❝❤ ❣✐↔✐ ❦❤→❝✱ ✤â ❧➔ ✤➦t p = x3 + 8✱
√
p + 3q = 10
q = 12 − x3 ✱ ✤÷❛ ✈➲ ❤➺
2
3
p + q = 20.
✶✳✾ P❤÷ì♥❣ ♣❤→♣ ❜➜t ✤➥♥❣ t❤ù❝
P❤÷ì♥❣ tr➻♥❤✱ ❤➺ ♣❤÷ì♥❣ tr➻♥❤✱ ❜➜t ♣❤÷ì♥❣ tr➻♥❤ ✈➔ ❜➜t ✤➥♥❣ t❤ù❝
❝â ♠è✐ ❧✐➯♥ ❤➺ ❝❤➦t ❝❤➩ ✈î✐ ♥❤❛✉✳ ❈❤➥♥❣ ❤↕♥ ❦❤✐ ❝❤ù♥❣ ♠✐♥❤ ♠ët ❜➜t
✤➥♥❣ t❤ù❝✱ t❛ ❝➛♥ ❞ü ✤♦→♥ ❞➜✉ ❜➡♥❣ ①↔② r❛ ❦❤✐ ♥➔♦✱ ✤✐➲✉ ♥➔② ❞➝♥ tî✐
t➻♠ ♠ët ♥❣❤✐➺♠ ♥➔♦ ✤â ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤✱ ❤➺ ♣❤÷ì♥❣ tr➻♥❤✳ ◗✉❛ ✤➙②
t❤➜② r➡♥❣ ✈✐➺❝ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤✱ ❤➺ ♣❤÷ì♥❣ tr➻♥❤ t❤ü❝ sü ❝â þ ♥❣❤➽❛✳
❇ð✐ ✈➟②✱ tr♦♥❣ q✉→ tr➻♥❤ ❣✐↔✐ t♦→♥ ❜➜t ✤➥♥❣ t❤ù❝ s➩ ♥↔② s✐♥❤ ♥❤✉ ❝➛✉
t➻♠ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤✱ ❤➺ ♣❤÷ì♥❣ tr➻♥❤✳ ◆❤✐➲✉ ❜➔✐ t♦→♥ ✈➲
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✸✾
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
♣❤÷ì♥❣ tr➻♥❤ ❤➺ ♣❤÷ì♥❣ tr➻♥❤ ❧↕✐ ❧➔ sü ❝❤❡ ❞➜✉ ❝õ❛ ♠ët ❜➜t ✤➥♥❣ t❤ù❝
♥➔♦ ✤â✳ ❉➜✉ ❤✐➺✉ ❝õ❛ ♥❤ú♥❣ ♣❤÷ì♥❣ tr➻♥❤ ♥➔② ❧➔ sè ♣❤÷ì♥❣ tr➻♥❤ ➼t
❤ì♥ sè ➞♥✱ ♣❤÷ì♥❣ tr➻♥❤ r➜t ♣❤ù❝ t↕♣✱ ❦❤æ♥❣ ♠➝✉ ♠ü❝✱ ♠❛♥❣ ❜â♥❣
❞→♥❣ ❝õ❛ ♠ët ❜➜t ✤➥♥❣ t❤ù❝ ♥➔♦ ✤â✳ ▼ët ✤✐➲✉ ✤➦❝ ❜✐➺t ❝õ❛ ♣❤÷ì♥❣
♣❤→♣ ♥➔② ❧➔ ♥➳✉ ✤♦→♥ ✤÷ñ❝ ♥❣❤✐➺♠ s➩ ❣â♣ ♣❤➛♥ r➜t ❧î♥ ✈➔♦ t❤➔♥❤
❝æ♥❣ ❝õ❛ ❧í✐ ❣✐↔✐✳
❚❛ ❝➛♥ ❧÷✉ þ ♠ët sè ❜➜t ✤➥♥❣ t❤ù❝ q✉❡♥ t❤✉ë❝ s❛✉✿
✶✳ |A| = | − A| ≥ 0✳ ❉➜✉ ✧❂✧ ①↔② r❛ ⇔ A = 0✳
✷✳ |A| ≥ A✳ ❉➜✉ ❜➡♥❣ ①↔② r❛ ⇔ A ≥ 0✳
✸✳ a2 ≥ 0, ∀a✳ ❉➜✉ ✧❂✧ ❝â ❦❤✐✿ a = 0✳
✹✳ |a| ≥ a, ∀a✳ ❉➜✉ ✧❂✧ ❝â ❦❤✐✿ a ≥ 0✳
✺✳ |a| + |b| ≥ |a + b|✳ ❉➜✉ ✧❂✧ ❝â ❦❤✐✿ ab ≥ 0✳
✻✳ |a| − |b| ≤ |a − b|✳ ❉➜✉ ✧❂✧ ❝â ❦❤✐✿ ab ≥ 0 ✳
|a| ≥ |b|
✼✳ a2 + b2 ≥ 2ab✳ ❉➜✉ ✧❂✧ ❝â ❦❤✐✿ a2 = b✳
✽✳ (a + b)2 ≥ 4ab ⇔ ab ≤ a +2 b ✳ ❉➜✉ ✧❂✧ ❝â ❦❤✐✿ a = −b✳
✾✳ a1 + 1b ≥ a +4 b (a; b > 0)✳ ❉➜✉ ✧❂✧ ❝â ❦❤✐✿ a = b✳
✶✵✳ ab + ab ≥ 2(ab > 0)✳ ❉➜✉ ✧❂✧ ❝â ❦❤✐✿ a = b✳
✶✶✳ ❇➜t ✤➥♥❣ t❤ù❝ ❈❛✉❝❤②✿
❱î✐ ♥ sè t❤ü❝ ❞÷ì♥❣✿ a1; a2; ...; an
❉↕♥❣ ✶✳ a1 + a2 +n ... + an ≥ √a1a2...an
n
❉↕♥❣ ✷✳ a1 + a2 + ... + an ≥ n √a1a2...an
❉↕♥❣ ✸✳ ( a1 + a2 +n ... + an )n ≥ a1a2...an
❉➜✉ ✧❂✧ ❝â ❦❤✐✿ a1 = a2 = ... = an
✶✷✳ ❇➜t ✤➥♥❣ t❤ù❝ ❇✉♥❤✐❛❦♦✈s❦②✿
n
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✹✵
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❱î✐ ✷ ❜ë sè t❤ü❝ ❜➜t ❦➻✿ a1; a2; ...; an✱ b1; b2; ...; bn✿
❉↕♥❣ ✶✳ (a1b1 + a2b2 + ... + anbn)2 ≤ (a21 + a22 + ... + a2n)(b21 + b22 + ... + b2n)
❉↕♥❣ ✷✳ |a1b1+a2b2+...+anbn| ≤ (a21 + a22 + ... + a2n)(b21 + b22 + ... + b2n)
❉➜✉ ✧❂✧ ❝â ❦❤✐✿ ab 1 = ab 2 = ... = ab n
1
2
n
❉↕♥❣ ✸✳ a1b1+a2b2+...+anbn ≤
(a21 + a22 + ... + a2n )(b21 + b22 + ... + b2n )
❉➜✉ ✧❂✧ ❝â ❦❤✐✿ ab 1 = ab 2 = ... = ab n > 0
1
2
n
✶✸✳ ❇➜t ✤➥♥❣ t❤ù❝ ❈❛✉❝❤②✲❙✇❛r❝❤③✿
❱î✐ ∀xi > 0; i = 1, n✱ t❛ ❝â✿
a21 a22
a2n
(a1 + a2 + ... + an )2
+
+ ... +
≥
x1 x2
xn
x1 + x2 + ... + xn
❈❤ù♥❣ ♠✐♥❤✿ ❳➨t
a1 √
a2 √
an √
(a1 + a2 + ... + an )2 = ( √ . x1 + √ . x2 + ... + √ . xn )2
x1
x2
xn
≤(
a2
a21 a22
+
+ ... + n )(x1 + x2 + ... + xn )
x1 x2
xn
✭⑩♣ ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❇✉♥❤✐❛❦♦✈s❦②✮
a1 √
a2 √
an √
(a1 + a2 + ... + an )2 = ( √ . x1 + √ . x2 + ... + √ . xn )2
x1
x2
xn
a21 a22
a2n
≤( +
+ ... + )(x1 + x2 + ... + xn )
x1 x2
xn
❱➟② ❜➜t ✤➥♥❣ t❤ù❝ ✤÷ñ❝ ❝❤ù♥❣ ♠✐♥❤✳
❚❛ ❝ò♥❣ ✤✐ t➻♠ ❤✐➸✉ ♣❤÷ì♥❣ ♣❤→♣ ♥➔② t❤æ♥❣ q✉❛ ❝→❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
x−2+
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
√
4 − x = x2 − 6x + 11
✹✶
(1)
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥
x−2≥0
4−x≥0
⇔ 2 ≤ x ≤ 4.
⑩♣ ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❇✉♥❤✐❛❦♦✈s❦② ❝❤♦ ✈➳ tr→✐ ✭✶✮ t❛ ❝â✿
√
√
VT = 1. x − 2 + 1. 4 − x ≤
(12 + 12 ) (x − 2 + 4 − x) = 2
(2)
❚❛ ❧↕✐ ❝â✿
VP = x2 − 6x + 11 = (x − 3)2 + 2 ≥ 2
(3)
▼➔ ✭✶✮ ⇔ ❱❚ ❂ ❱P✳
❙✉②r❛ t❛ ❝â ✭✶✮ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐ ❞➜✉ ✧❂✧ ð ✭✷✮ ✈➔ ✭✸✮ ①↔② r❛
x−2 = 4−x
1
1
⇔
⇔ x = 3.
x=3
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ ❞✉② ♥❤➜t x = 3✳
❇➔✐ t♦→♥ ✷✳ ✭✣➲ ♥❣❤à ❖❧②♠♣✐❝ ✸✵✴✵✹✴✷✵✶✶✮✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
x3000 + 500x3 + 1500x + 1999 = 0.
✭✶✮
●✐↔✐✳ ⑩♣ ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❈❛✉❝❤② ❝❤♦ x3000 ✈➔ 2999 sè ✶✱ t❛ ✤÷ñ❝
√
✭✷✮
❉➜✉ ❜➡♥❣ tr♦♥❣ ✭✷✮ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐ x = −1✳ ❚÷ì♥❣ tü✿
√
x3000 + 999 ≥ 1000
x3000 = 1000 x3 ≥ −1000x3 . ✭✸✮
❉➜✉ ❜➡♥❣ tr♦♥❣ ✭✸✮ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐ x = −1✳ ❚ø ✭✷✮ ✈➔ ✭✸✮✱ t❛
✤÷ñ❝
x3000 + 2999 ≥ 3000
3000
x3000 = 3000 |x| ≥ −3000x.
1000
2x3000 + 3998 ≥ −3000x − 1000x3
⇔ x3000 + 500x3 + 1500x + 1999 ≥ 0✭✹✮
▼➔ ✭✶✮ ♥❣❤➽❛ ❧➔ ❞➜✉ ❜➡♥❣ ð ✭✹✮ ①↔② r❛✱ tù❝ ❧➔ ❞➜✉ ❜➡♥❣ ✭✷✮ ✈➔ ✭✸✮ ✤ç♥❣
t❤í✐ ①↔② r❛✳ ❱➟② (1) ⇔ x = −1✳ P❤÷ì♥❣ tr➻♥❤ ✭✶✮ ❝â ♥❣❤✐➺♠ ❞✉② ♥❤➜t
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✹✷
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
x = −1✳
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
2(
x10 y 10
16
16
2 2 2
+
)
+
x
+
y
=
4(1
+
x
y ) − 10
y2
x2
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x, y = 0
⑩♣ ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❈æ✲s✐ ❝❤♦ ❜è♥ sè ❞÷ì♥❣ t❛ ❝â✿
10 10
.y
x10 y 10
4 x
+ 2 + 1 + 1 ≥ 4.
= 4x2 y 2
2
2
y
x
x .y 2
x10 y 10
⇒ 2( 2 + 2 + 2) ≥ 8x2 y 2
y
x
❱➔✿
x16 + y 16 + 1 + 1 ≥ 4. 4 x16 .y 16 = 4x4 y 4
x10 y 10
⇒ 2( 2 + 2 + 2) + x16 + y 16 + 2 ≥ 4x4 y 4 + 8x2 y 2
y
x
x10 y 10
2
⇒ 2( 2 + 2 ) + x16 + y 16 ≥ 4(1 + x2 y 2 ) − 10
y
x
❉➜✉ ✤➥♥❣ t❤ù❝ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐✿
x2 = y 2 = 1 ⇔ |x| = |y| = 1
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ❝→❝ ♥❣❤✐➺♠ (x; y) ❧➔✿ ✭✶❀✶✮✱✭✲✶❀✲✶✮✱
✭✶❀✲✶✮✱✭✲✶❀✶✮✳
❇➔✐ t♦→♥ ✹✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ x4 + 4 = 2√x4 + 4 + 2√x4 − 4✳
●✐↔✐✳ ⑩♣ ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❈❛✉❝❤② ❝❤♦ ❤❛✐ sè ❦❤æ♥❣ ➙♠ t❛ ❝â✿
x4 + 4 ≥ 4x2
✭✶✮
❚❤❡♦ ❇✣❚ ❈❛✉❝❤②✲s❝❤✇❛r③ t❛ ❝â✿
(a + b)2 ≤ 2(a2 + b2 ) ⇔ a + b ≤
❉➜✉ ✧❂✧ ①↔② r❛ ❦❤✐ a = b ≥ 0✳
❚rð ❧↕✐ ❜➔✐ t♦→♥✱ →♣ ❞ö♥❣ ❇✣❚ tr➯♥ t❛ ❝â ✿
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✹✸
2(a2 + b2 )
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
√
√
2 x4 + 4 + 2 x4 − 4 ≤ 2[4(x4 + 4) + 4(x4 − 4)]
√
√
⇔ 2 x4 + 4 + 2 x4 − 4 ≤ 4x2
❚ø ✭✶✮ ✈➔ ✭✷✮ t❛ ❝â ❞➜✉ ❜➡♥❣ ①↔② r❛ ❦❤✐✿
x4 = 4
√
√
2 x4 + 4 = 2 x4 − 4
✭❤➺ ✈æ ♥❣❤✐➺♠✮
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ✈æ ♥❣❤✐➺♠✳
❇➔✐ t♦→♥ ✺✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
2−
x2
+
4
2−
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥
4
=4−
x2
x 2
+
2 x
(1)
2
2− x ≥0
x2 ≤ 8
4
⇔
x2 ≥ 2
2− 4 ≥0
2
x √
√
√ √
⇔ x ∈ −2 2; − 2 ∪
2; 2 2
✣➦t t = x2 ✱ ♣❤÷ì♥❣ tr➻♥❤ ✭✶✮ trð t❤➔♥❤
2 − t2 + 2 −
1
1
=
4−
t
+
t2
t
1
1
⇔ t+ 2 − t2 + + 2 − 2 = 4 (2)
t
t
❙û ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❇✉♥❤✐❛❦♦✈s❦② t❛ ❝â✿
t+ 2 − t2 = 1.t+1. 2 − t2 ≤
(12 + 12 ) (t2 + 2 − t2 ) = 2
1
1
1
+ 2 − 2 = 1.t+1. 2 − 2 ≤
t
t
t
(12 + 12 )
1
1
+
2
−
t2
t2
(3)
= 2 (4)
❚ø ✭✸✮ ✈➔ ✭✹✮ t❛ s✉② r❛
1
≤ 4 (5)
t2
√
❚❛ ❝â ✭✷✮ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐ ❞➜✉ ❜➡♥❣ ð ✭✺✮ ①↔② r❛ ⇔ t = 2 − t2 ⇔ t = 1✳
❙✉② r❛ t❛ ❝â ✭✶✮ ⇔ x2 = 1 ⇔ x = 2.
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ ❞✉② ♥❤➜t x = 2✳
t+
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
2 − t2 +
1
+
t
✹✹
2−
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❈❤÷ì♥❣ ✷
▼ët sè ❝→❝❤ s→♥❣ t↕♦ r❛ ✤➲ t♦→♥
♠î✐
✷✳✶ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➺ sè ❜➜t
✤à♥❤
❚❛ s➩ s→♥❣ t→❝ ❝→❝ ♣❤÷ì♥❣ tr➻♥❤ ✤÷ñ❝ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➺ sè
❜➜t ✤à♥❤ ♥❤÷ s❛✉✿
❱➼ ❞ö ✶✳ ❚❛ ❝â✿
(a − b + 1) (2a − b + 3) = 0 ⇔ 2a2 + b2 − 3ab + 5a − 4b + 3 = 0. ❚ø
√
√
✤➙② ❧➜② a = 1 + x ✈➔ b = 1 − x t❛ ✤÷ñ❝
√
√
√
2x + 2 + 1 − x − 3 1 − x2 + 5 1 + x − 4 1 − x + 3 = 0.
❘ót ❣å♥ t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
√
4 1 − x = x + 6 − 3 1 − x2 + 5 1 + x.
(1)
●✐↔✐✳ ❚➟♣ ①→❝ ✤à♥❤ −1 ≤ x ≤ 1✳
P❤÷ì♥❣ tr➻♥❤ ✭✶✮ t÷ì♥❣ ✤÷ì♥❣ ✈î✐
√
√
√
2(1 + x) + (1 − x) − 3 1 − x2 + 5 1 + x − 4 1 − x = 0
√
√
✣➦t u = 1 + x ✈➔ v = 1 − x✱ t❛ ✤÷ñ❝
✹✺
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
2u2 + v 2 − 3uv + 5u − 4v + 3 = 0 ⇔ (u − v + 1) (2u − v + 3) = 0
u − v + 1 = 0 (1)
⇔
2u − v + 3 = 0 (2)
√
√
• ◆➳✉ u − v + 1 = 0 ⇒ 1 + x = 1 − x − 1
√
√
√
2
⇔ 1+x+1= 1−x⇔
1+x+1 =1−x
√
⇔1+x+1+2 1+x=1−x
√
⇔
2 1 + x = −1 − 2x
4 (1 + x) = 1 + 4x + 4x2
⇔
x ≤ −1
2
√
2
4x = 3
− 3
⇔
⇔x=
✳
x ≤ −1
2
2
• ◆➳✉ 2u − v + 3 = 0
√
√
√
√
⇒2 1+x− 1−x+3=0⇔2 1+x+3= 1−x
√
2
⇔ 2 1+x+3 =1−x
√
⇔ 4 (1 + x) + 9 + 12 1 + x = 1 − x
√
⇔
12 1 + x = −5x − 12
144(1 + x) = 25x2 + 120x + 144
⇔
x ≤ −12
5
x=0
√
2
25x − 24x = 0
2 6
⇔
⇔
✳
x=
x ≤ −12
5
−12
5
x≤
5
❙✉② r❛ ❦❤æ♥❣ ❝â ❣✐→ trà x t❤ä❛ ♠➣♥✳
√
− 3
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ♠ët ♥❣❤✐➺♠ ❞✉② ♥❤➜t x = 2 ✳
❱➼ ❞ö ✷✳ ❳✉➜t ♣❤→t tø ✤➥♥❣ t❤ù❝
(u − 2v)2 − 9 = 0 ⇔ u2 + 4v 2 − 9 − 4uv = 0
1
9
⇔ u2 + v 2 − = uv
4
4
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✹✻
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✣➦t u = 6x + 1✱ v =
√
x2 + 3 ✳
❚❛ ❝â✿
9
1
(6x + 1)2 + (x2 + 3) − = (6x + 1)
4
4
x2 + 3.
❘ót ❣å♥ t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✿
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ 10x2 + 3x + 1√= (6x + 1) √x2 + 3
✣→♣ sè✳ P❤÷ì♥❣ tr➻♥❤ ❝â t➟♣ ♥❣❤✐➺♠ S = 1; 74− 3 ✳
✷✳✷ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ✤÷❛ ✈➲
❤➺
❱➼ ❞ö ✶✳ ❳➨t ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❤❛✐ ❝â ❝↔ ❤❛✐ ♥❣❤✐➺♠ ❧➔ sè ✈æ t➾
5x2 − 2x − 1 = 0 ⇔ 2x = 5x2 − 1
❉♦ ✤â ①➨t
2y = 5x2 − 1
5x2 − 1
⇒ 2x = 5
2
2x = 5y 2 − 1
2
−1
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ 8x − 5 5x2 − 1 2 = −4.
●✐↔✐✳ ✣➦t 2y = 5x2 − 1✳ ❑❤✐ ✤â
2y = 5x2 − 1
8x − 5.4y 2 = −4
⇔
2y = 5x2 − 1 (1)
2x = 5y 2 − 1 (2)
▲➜② ✭✶✮ trø ✭✷✮ t❤❡♦ ✈➳ t❛ ✤÷ñ❝✿
2
2 (y − x) = 5 x − y
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
2
⇔
y−x=0
2 = −5 (x + y)
✹✼
⇔
y=x
y=−
5x + 2
.
5
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
√
1
±
6
5x2 − 2x − 1 = 0 ⇔ x =
.
5
❱î✐ y = x✱ t❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝
5x + 2
• ❱î✐ y = −
✱ t❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝
5
•
√
10x + 4
−5 ± 50
2
2
−
= 5x − 1 ⇔ 25x + 10x − 1 = 0 ⇔ x =
.
5
25
❱➟②
√
√
1 ± 6 −5 ± 50
♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ❜è♥ ♥❣❤✐➺♠ 5 ✱ 25 ✳
√
√
❚❛ ❝â 4x3 − 3x = 23 ⇔ 6x = 8x3 − 3✳ ❱➟② ①➨t
√
√
√
6y = 8x3 − 3
8x3 − 3
⇒ 6x = 8
− 3
√
6
6x = 8y 3 − 3
❱➼ ❞ö ✷✳
√
√
⇒ 1296x + 216 3 = 8 8x3 − 3
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
3
√
√ 3
⇒ 162x + 27 3 = 8x3 − 3 .
√
√ 3
162x + 27 3 = 8x3 − 3 .
●✐↔✐✳ ✣➦t 6y = 8x3 − √3✳ ❚❛ ❝â ❤➺
√
6y = 8x3 − 3
⇔
√
162x + 27 3 = 216y 3
6y = 8x3 −
6x = 8y 3 −
√
√
3 (1)
3 (2)
▲➜② ✭✶✮ trø ✭✷✮ t❤❡♦ ✈➳ t❛ ✤÷ñ❝
6(y − x) = 8(x3 − y 3 ) ⇔ (x − y) 8 x2 + xy + y 2 + 6 = 0. (3)
❱➻ x2 + xy + y2 ≥ 0 ♥➯♥ 8
t❛ ✤÷ñ❝ x = y✳
❚❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝
6x = 8x3 −
√
x2 + xy + y 2 ≥ 0 + 6 > 0✳
√
3
π
⇔ 4x3 − 3x = cos
2
6
α
α
cosα= 4cos3 − 3 cos ✱ t❛ ❝â
3
3
π
α
α
cos = 4cos3 − 3 cos ,
6
18
18
3 ⇔ 4x3 − 3x =
❙û ❞ö♥❣ ❝æ♥❣ t❤ù❝
❉♦ ✤â tø ✭✸✮
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✹✽
(4).
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
11α
11α
11π
= 4cos3
− 3 cos
,
6
18
18
13π
13α
13α
cos
= 4cos3
− 3 cos
.
6
18
18
π
11π
13π
x = cos ✱ x = cos
✱
x = cos
❧➔ t➜t ❝↔
18
18
6
cos
❱➟②
❝→❝ ♥❣❤✐➺♠ ❝õ❛
♣❤÷ì♥❣ tr➻♥❤ ✭✹✮✱ ✈➔ ❝ô♥❣ ❧➔ t➜t ❝↔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✤➣
❝❤♦✳
▲÷✉ þ✳ P❤➨♣ ✤➦t 6y = 8x3 − √3 ✤÷ñ❝ t➻♠ r❛ ♥❤÷ s❛✉✿ ❚❛ ✤➦t
√
ax + b = 8x3 − 3✭ ✈î✐ ❛✱ ❜ s➩ t➻♠ s❛✉✮✳
❑➳t ❤ñ♣ ✈î✐ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ❤➺
√
ax + b = 8x3 − 3
√
162x + 27 3 = a3 y 3 + 3a2 by 2 3ab2 y + b3 .
❈➛♥ ❝❤å♥ a ✈➔ b s❛♦ ❝❤♦✿
❱➟② t❛ ❝â
√
a
3
8
b
+
= 3= √
162 a
27 3 − b3 ⇒
3a2 b = 3ab2 = 0
√
♣❤➨♣ ✤➦t 6x = 8x3 − 3✳
b=0
a=6
❱➼ ❞ö ✸✳❳✉➜t ♣❤→t tø (x + 1) = (x + 2)2✳ ❱➟② t❛ ①➨t ❤➺
y + 1 = (x + 2)2
x + 1 = (y + 2)2
❚ø ❤➺ tr➯♥✱ sû ❞ö♥❣ ♣❤÷ì♥❣ ♣❤→♣ t❤➳ t❛ ✤÷ñ❝
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
√
x + 1 = x2 + 4x + 5
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ √x + 1 = x2 + 4x + 5
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x + 1 ≥ 0 ⇔ x ≥√ −1✳
❱✐➳t ❧↕✐ ♣❤÷ì♥❣ tr➻♥❤ ❞÷î✐ ❞↕♥❣
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
x + 1 = (x + 2)2 + 1
✹✾
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
√
✣➦t y + 2 = x + 1✱ ✤✐➲✉ ❦✐➺♥ y + 2 ≥ 0 ⇔ y ≥ −2✳
❑❤✐ ✤â✱ ♣❤÷ì♥❣ tr➻♥❤ ✤÷ñ❝ ❝❤✉②➸♥ t❤➔♥❤ ❤➺✿
y + 2 = (x + 2)2 + 1
⇔
(y + 2)2 = x + 1
✭✯✮
y + 1 = (x + 2)2
(1)
x + 1 = (y + 2)2
(2)
⇒ x − y = − (x + 2)2 − (y + 2)2 ⇔ x − y = − (x − y) (x + y + 4)
✭❞♦ ✭✯✮ ✈➔ x > 0)
❱î✐ x = y t❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝ x2 + 3x + 3 = 0 ✭✈æ ♥❣❤✐➺♠✮
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ✈æ ♥❣❤✐➺♠✳
⇔ (x − y) (x + y + 5) = 0 ⇔ x = y
❱➼ ❞ö ✹✳ ❈❤å♥ ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❝❤➾ ❝â ❤❛✐ ♥❣❤✐➺♠ ❧➔ ✵ ✈➔ ✶ ❧➔
❚ø ♣❤÷ì♥❣ tr➻♥❤ ♥➔② t❛ t❤✐➳t ❧➟♣ ♠ët ❤➺ ✤è✐ ①ù♥❣ ❧♦↕✐
❤❛✐✱ s❛✉ ✤â ❧↕✐ q✉❛② ✈➲ ♣❤÷ì♥❣ tr➻♥❤ ♥❤÷ s❛✉✿
11x = 10x + 1.
11x = 10y + 1
11y = 10x + 1
⇒
y = log11 (10x + 1)
11x = 10y + 1
11x − 1
= log11 (10x + 1) .
10
11x = 10log11 (10x + 1) + 1 ⇒ 11x = 2log11 (10x + 1)5 + 1.
⇒
❙✉② r❛
❝â ❜➔✐ t♦→♥ s❛✉
❚❛
❇➔✐ t♦→♥ ✹✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ 11x = 2log11(10x + 1)5 + 1.
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x > −1
✳ ✣➦t y = log11 (10x + 1)✱ ❦❤✐ ✤â
10
11y = 10x + 1✳
❑➳t ❤ñ♣ ✈î✐ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦✱ t❛ ❝â ❤➺
11x = 10y + 1 (1)
11y = 10x + 1 (2)
▲➜② ✭✶✮ trø ✭✷✮ t❤❡♦ ✈➳ t❛ ✤÷ñ❝
11x − 11y = 10y − 10x ⇔ 11x + 10x = 11y + 10y
(3)
❳➨t ❤➔♠ sè f (t) = 11t + 10t✳ ❚❛ ❝â f (t) = 11t ln 11 + 10 > 0, ∀t ∈ R.
❱➟② ❤➔♠ sè f ✤ç♥❣ ❜✐➳♥ tr➯♥ R✳ ▼➔ ✭✸✮ ❝❤➼♥❤ ❧➔ f (x) = f (y) ♥➯♥
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✺✵
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
x = y✳
❚❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝ 11x = 10x + 1 ⇔ 11x − 10x − 1 = 0 (4)
; +∞ ✳ ❚❛ ❝â
❳➨t ❤➔♠ sè g (x) = 11x − 10x − 1 tr➯♥ ❦❤♦↔♥❣ −1
10
g (x) = 11x ln 11 − 10
g (x) = 11x (ln 11)2 > 0.
; +∞ ✱ s✉② r❛ ✤ç
❱➟② ❤➔♠ sè g ❝â ✤ç t❤à ❧✉æ♥ ❧ã♠ tr➯♥ ❦❤♦↔♥❣ −1
10
t❤à ❝õ❛ ❤➔♠ g ✈➔ trö❝ ❤♦➔♥❤ ❝â ✈î✐ ♥❤❛✉ ❦❤æ♥❣ q✉→ ❤❛✐ ✤✐➸♠ ❝❤✉♥❣✱
s✉② r❛ ✭✹✮ ❝â ❦❤æ♥❣ q✉→ ❤❛✐ ♥❣❤✐➺♠✳ ▼➔ g(1) = 0✱ g(0) = 0 ♥➯♥ x = 0
✈➔ x = 1 ❧➔ t➜t ❝↔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ✭✹✮✳
◆❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❧➔ x = 0 ✈➔ x = 1✳
✷✳✸ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➔♠ sè
❱➼ ❞ö ✶✳ ❳➨t ❤➔♠ sè f (x) = √4x − 1 + √4x2 − 1✳ ❍➔♠ f (x) ①→❝
✤à♥❤ tr➯♥ D =
1
; +∞
2
f (x) = √
✱ ❝â ✤↕♦ ❤➔♠
2
4x
+√
> 0, ∀x ∈ D.
4x − 1
4x2 − 1
❱➟② ❤➔♠ f (x) ❧✉æ♥ ✤ç♥❣ ❜✐➳♥ tr➯♥
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥
√
1
; +∞
2
4x − 1 +
✳ ❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
√
4x2 − 1 = 1
✭✶✮
4x − 1 ≥ 0
1
⇔x≥ .
2
4x2 − 1 ≥ 0
◆❤➟♥ ①➨t r➡♥❣✿ sè ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✭✶✮ ❧➔ sè ❣✐❛♦ ✤✐➸♠ ❝õ❛ ✤ç
√
√
t❤à ❤➔♠ sè y = f (x) = 4x − 1 + 4x2 − 1 ✈➔ ✤÷í♥❣ t❤➥♥❣ y = 1✳
√
√
❳➨t ❤➔♠ sè f (x) = 4x − 1 + 4x2 − 1✿
1
• ▼✐➲♥ ①→❝ ✤à♥❤ D = ; +∞
2
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✺✶
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
•
✣↕♦ ❤➔♠✿
f (x) = √
4x
2
+√
> 0, ∀x ∈ D.
4x − 1
4x2 − 1
❉♦ ✤â ❤➔♠ sè ❧✉æ♥ ✤ç♥❣ ❜✐➳♥✱ ♥➯♥ ♥➳✉ ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ t❤➻ ✤â
❧➔ ❞✉② ♥❤➜t✳
❚❛ t❤➜② x = 12 t❤ä❛ ♠➣♥ ♣❤÷ì♥❣ tr➻♥❤✳
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ ❞✉② ♥❤➜t x = 21 ✳
❱➼ ❞ö ✷✳ ❳✉➜t ♣❤→t tø ❤➔♠ sè f (x) =
❤➔♠
√
x+
x2 − x + 1 + x
❝â ✤↕♦
√
√
x2 − x + 1
2 x2 − x + 1 + 2x − 1
=
> 0, ∀x
f (x) =
√
√
√
2
2
2
2 x+ x −x+1 4 x+ x −x+1 x −x+1
x+
❚ø ♣❤÷ì♥❣ tr➻♥❤ ❤➔♠ f (x) = f (x + 1) t❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
x2 − x + 1 −
x+
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥
√
x2 − x + 1 ≥ 0
⇔
√
x + 1 + x2 + x + 1 ≥ 0
x+
x2 − x + 1 = 1
x+1+
√
√
(1)
x2 − x + 1 ≥ −x
(2)
x2 + x + 1 ≥ −x − 1
(3)
❚❛ ❝â✿
(2) ⇔
(3) ⇔
−x ≤ 0
x2 − x + 1 ≥ 0
−x ≥ 0
⇔
x≥0
x≤0
⇔ ∀x.
x2 − x + 1 ≥ x2
−1 − x ≤ 0
x2 + x + 1 ≥ 0
⇔
−1 − x ≥ 0
x ≥ −1
x ≤ −1
⇔ ∀x.
x2 + x + 1 ≥ (−1 − x)2
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✺✷
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❱➟② t❛ ✤÷ñ❝ D = R✳
❱✐➳t ❧↕✐ ♣❤÷ì♥❣ tr➻♥❤ ❞÷î✐ ❞↕♥❣✿
x+
•
x2 − x + 1 + x =
✣↕♦ ❤➔♠
x+1+
x2 − x + 1 + x + 1
(4)
√
√
x2 − x + 1
2 x2 − x + 1 + 2x − 1
f (x) =
=
√
√
√
2 x + x2 − x + 1 4 x + x2 − x + 1 x2 − x + 1
x+
◆❤➟♥ ①➨t✿
√
2 x2 − x + 1 + 2x − 1 =
(2x − 1)2 + 3 + 2x − 1
> |2x − 1| + 2x − 1 ≥ 0
⇒ f (x) > 0, ∀x
❙✉② r❛ ❤➔♠ sè ❧✉æ♥ ✤ç♥❣ ❜✐➳♥✳
❑❤✐ ✤â✿ (4) ⇔ f (x) = f (x + 1) ⇔ x = x + 1 ✭✈æ ♥❣❤✐➺♠✮✳
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ✈æ ♥❣❤✐➺♠✳
❱➼ ❞ö ✸✳ ❳✉➜t ♣❤→t tø ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❝â ❝→❝❤ ❣✐↔✐ r➜t ❝ì ❜↔♥✱ ✤â
❧➔ 7x−1 = 6x − 5✳ ❳➨t ♠ët ❤➔♠ sè ✤ì♥ ✤✐➺✉ φ (t) = t + 6log7t✳ ❑❤✐ ✤â
φ 7x−1 = φ (6x − 5)
⇔ 7x−1 + 6log7 7x−1 = (6x − 5) + 6log7 (6x − 5)
⇔ 7x−1 + 6(x − 1) = (6x − 5) + 6log7 (6x − 5)
⇔ 7x−1 = 1 + 2 log (6x − 5)3 .
❚❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x > 56 ✳ ❚❛ ❝â
7x−1 = 1 + 2 log (6x − 5)3
(1)✳
(1) ⇔ 7x−1 − 6log7 (6x − 5) = −6 (x − 1) + (6x − 5) (2)
⇔ 7x−1 + 6 (x − 1) = (6x − 5) + 6log7 (6x − 5)
⇔ φ 7x−1 = φ (6x − 5)✱
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✈î✐ φ (t) = t + 6log7t (3)
✺✸
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❍➔♠ sè φ ✤ç♥❣ ❜✐➳♥ tr➯♥ (0; +∞) ✈➻ φ (t) = 1 + t ln6 7
❱➟②
(3) ⇔ 7x−1 = 6x − 5 ⇔ 7x−1 − 6x + 5 = 0
> 0, ∀t > 0✳
(4)
❈→❝❤ ✶✳ ❉➵ t❤➜② r➡♥❣ x = 1✱ x = 2 t❤ä❛ ✭✹✮✳ ❳➨t ❤➔♠ f (x) =
7x−1 − 6x + 5
tr➯♥ R✳ ❚❛ ❝â
f (x) = 7x−1 ln 7 − 6❀ f (x) = 7x−1 (ln 7)2 > 0, ∀x ∈ R
❱➟② ❤➔♠ f ❝â ✤ç t❤à ❧✉æ♥ ❧✉æ♥ ❧ã♠ ♥➯♥ ❝➢t trö❝ ❖① t↕✐ ❦❤æ♥❣ q✉→ ❤❛✐
✤✐➸♠✱ s✉② r❛ ✭✹✮❝â ❦❤æ♥❣ q✉→ ❤❛✐ ♥❣❤✐➺♠✳ ❱➟② x = 1✱ x = 2 ❧➔ t➜t ❝↔
❝→❝ ♥❣❤✐➺♠ ❝õ❛ ✭✹✮✳ P❤÷ì♥❣ tr➻♥❤ ✭✶✮ ❝â t➟♣ ♥❣❤✐➺♠ ❧➔ S = {1, 2}✳
❈→❝❤ ✷✳ ❚❛ ❝â f (x) = 0 ⇔ 7x−1 = ln67 ⇔ x = x0 = 1+log7 (6.log7e)✳
❱➻ ✈î✐ ♠å✐ x ∈ R t❤➻ f (x) > 0 ♥➯♥ s✉② r❛ f ❧➔ ❤➔♠ ✤ç♥❣ ❜✐➳♥ tr➯♥
❘ ✈➔
f (x) < 0, ∀x ∈ (−∞; x0 ) ; f (x) > 0, ∀x ∈ (x0 ; +∞) .
❱➟② ❤➔♠ f ♥❣❤à❝❤ ❜✐➳♥ tr➯♥ (−∞; x0) ✈➔ ✤ç♥❣ ❜✐➳♥ tr➯♥ (x0; +∞)✱
❞♦ ✤â ✭✹✮ ❝â ❦❤æ♥❣ q✉→ ❤❛✐ ♥❣❤✐➺♠✳ ❱➟② x = 1✱ x = 2 ❧➔ t➜t ❝↔ ❝→❝
♥❣❤✐➺♠ ❝õ❛ ✭✹✮✳ P❤÷ì♥❣ tr➻♥❤ ✭✶✮ ❝â t➟♣ ♥❣❤✐➺♠ ❧➔ S = {1, 2}✳
▲÷✉ þ✳ P❤➨♣ ♣❤➙♥ t➼❝❤ ✭✷✮ ✤÷ñ❝ t➻♠ r❛ ♥❤÷ s❛✉✿ ❈➛♥ ❝❤å♥ α, β, γ
s❛♦ ❝❤♦
α + 6β = 0
1 = α (x − 1) + β (6x − 5) ⇒
−α − 5β = 1
⇔
α = −6
β = 1.
❱➼ ❞ö ✹✳ ❳➨t ❤➔♠ sè ✤ç♥❣ ❜✐➳♥ tr➯♥ R ❧➔ f (t) = 5t + t✳ ❚ø ♣❤÷ì♥❣
tr➻♥❤ ❤➔♠ f
5
1
3x
1
3x
= f (x − 1)✱
t❛ ❝â
√
1
3
x−1
+
=5
+x−1⇔
5
3x
1
x
√
3
1−x
⇔ 3x
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
5
1
x
1
− 3x
5
✺✹
−
1
5
1−x
3xx − 3x − 1
=
3x
= 3xx − 3x − 1
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✱
❇➔✐ t♦→♥ ✹✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ 3x
√
3
5
1
x
1
− 3x
5
✣→♣ sè✳ P❤÷ì♥❣ tr➻♥❤ ❝â ❤❛✐ ♥❣❤✐➺♠ ❧➔
x=
3+
√
21
6
,x =
3−
√
21
6
1−x
= 3xx − 3x − 1
.
✷✳✹ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❤➡♥❣ sè
❜✐➳♥ t❤✐➯♥
❱➼ ❞ö ✶✳ ❇➡♥❣ ❦❤❛✐ tr✐➸♥ ♣❤÷ì♥❣ tr➻♥❤ t➼❝❤
x2 − 3x − 1 − a
x2 − 5x − a = 0
t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
x4 − 8x3 + 2 (7 − a) x2 + (5 + 8a)x + a2 + a = 0.
(1)
●✐↔✐✳ ❚❛ ❝♦✐ ✭✶✮ ❧➔ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❤❛✐ ➞♥ a✿
a2 − 2x2 − 8x − 1 a + x4 − 8x3 + 14x2 + 5x = 0
❝â ∆ =
2x2 − 8x − 1
2
(2)
− 4 x4 − 8x3 + 14x2 + 5x = (2x − 1)2 ✳
2x2 − 8x − 1 + |2x − 1|
a=
2
⇔
(2) ⇔
2x2 − 8x − 1 − |2x − 1|
a=
2
❱➟②
x2 − 3x − 1 − a = 0 (3)
x2 − 5x − a = 0.
(4)
P❤÷ì♥❣ tr➻♥❤ ✭✸✮ ✈➔ ✭✹✮ ❧➛♥ ❧÷ñt ❝â ❜✐➺t t❤ù❝
∆1 = 9 + 4(1 + a) = 13 + 4a, ∆2 = 25 + 4a.
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✺✺
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
•
t❤➻ ✭✸✮ ✈æ ♥❣❤✐➺♠✳
◆➳✉ ∆1 < 0 ❤❛② a < −13
4
◆➳✉ ∆1 ≥ 0 ❤❛②
• ◆➳✉ ∆2 < 0 ❤❛②
•
−13
a≥
4
−25
a<
4
−25
a≥
4
t❤➻ (3) ⇔ x = x1,2 =
t❤➻ ✭✹✮ ✈æ ♥❣❤✐➺♠✳
3±
5±
√
13 + 4a
✳
2
√
25 + 4a
◆➳✉ ∆2 ≥ 0 ❤❛②
t❤➻ (4) ⇔ x = x3,4 =
✳
2
❚❛ ❝â ❦➳t ❧✉➟♥✿
✲ ◆➳✉ a < −25
t❤➻ ✭✶✮ ✈æ ♥❣❤✐➺♠✳
4
−13
✲ ◆➳✉ −25
≤a<
t❤➻ ✭✶✮ ❝â ❤❛✐ ♥❣❤✐➺♠ x = x3✱ x = x4✳
4
4
✲ ◆➳✉ a ≥ −13
t❤➻ ✭✶✮ ❝â ❜è♥ ♥❣❤✐➺♠ x = x1, x = x2,x = x3, x = x4.
4
•
❱➼ ❞ö ✷✳ ❳➨t a1 = x x−2 3 ✱ a2 = −2x2
4
4
4
a1 + a2 = x − 3 − 2x2 = − x + 3
x2
x2
⇒
4
a1 a2 = x − 3 −2x2 = −2 x4 − 3 .
x2
❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ➞♥ a s❛✉
❑❤✐ ✤â a1✱ a2 ❧➔ ♥❣❤✐➺♠
2 x4 − 3
x4 + 3
a2 x4 + 3
4
a +
a−2 x −3 =0⇔
+
a−
=0
x2
x
x3
x
2
6
a2 + 6
a2
3a
3a
3
3
+ ax + 3 − 2x + = 0 ⇔ 3 − 2x +
+ ax = 0.
⇔
x
x
x
x
x
√
❈❤♦ a = 2 19 t❛ ✤÷ñ❝
√
√
√
3 19
25
171
25 √
3
3
−
2x
+
+
19x
=
0
⇔
−
2x
+
+ 19x = 0.
x3
x
x2
x
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
√
25 √
3
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ x171
−
2x
+
+ 19x = 0. ✭✶✮
2
x
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥✿ x = 0.
• x ❧➔ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✭✶✮ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐
√
3 19
19 + 6 √
3
−
2x
+
+ 19x = 0.
x3
x
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✺✻
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❙✉② r❛
√
19
❧➔ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ s❛✉ ✈î✐ ➞♥ ❧➔ a✿
a2 + 6
3a
3
−2x +
+ax = 0 ⇔ x2 a2 + x4 + 3 a−2x6 +6x2 = 0
3
x
x
❚❛ ❝â ∆ = x8 + 6x4 + 9 − 4x2
6x2 − 2x6 = 9 x4 − 1
2
(2)
✳ ❱➻ ✈➟②
−x4 − 3 − 3 x4 − 1
= −2x2
(3)
a=
2
2x
(2) ⇔
−x4 − 3 + 3 x4 − 1
x4 − 3
=
(4)
a=
2x2
x2
√
√
a = 19 ✈➔♦ ✭✸✮ t❛ ✤÷ñ❝ 19 = −2x2 ✳ P❤÷ì♥❣ tr➻♥❤
❚❤❛②
♥❣❤✐➺♠✳
√
• ❚❤❛② a = 19 ✈➔♦ ✭✹✮ t❛ ✤÷ñ❝
•
♥➔② ✈æ
√
√
√
x4 − 3
19
+
31
4
2
2
19 =
⇔
x
−
19x
−
3
=
0
⇔
x
=
.
x2
2
√
√
19 + 31
♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ❤❛✐ ♥❣❤✐➺♠ ❧➔ ±
.
2
√
❱➟②
❱➼ ❞ö ✸✳ ❳➨t
a1 = t2 − 2t✱ a2 = −t2 + 4t ⇒
❱➟② a1 ✈➔ a2 ❧➔ ♥❣❤✐➺♠ ❝õ❛
a1 + a2 = 2t
a1 a2 = −t4 + 6t3 − 8t2 .
♣❤÷ì♥❣ tr➻♥❤ ➞♥ a s❛✉
a2 − 2ta − t4 + 6t3 − 8t2 = 0.
❈❤å♥ a = 4 t❛ ✤÷ñ❝
16 − 8t − t4 + 6t3 − 8t2 = 0 ⇔ t4 − 6t3 + 8t2 + 8t − 16 = 0
16 8
⇔ t2 − 6t + 8 = 2 − ✳
t
t
▲➜② t = log3x✱ t❛ ✤÷ñ❝
log23 x − 6log3 x + 8 = 16 log2x 3 − 8logx 3
⇔ log23 x − 6log3 x + log3 9 + 6 = 16 log2x 3 − 4.2logx 3
x
⇔ log23 x − 6log3 √
+ 6 = 16 log2x 3 − 4logx 9.
3
3
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✺✼
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
x
+ 6 = 16 log2x 3 − 4logx 9
log23 x − 6log3 √
3
3
(1)
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ 0 < x = 1✳ ✣➦t log3x = t✳ ❑❤✐ ✤â
1
2
x
1
1
logx 3 = , logx 9 = , log3 √
= log3 x − log3 3 3 = t − .
3
t
t
3
3
❚❤❛② ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✭✶✮ t❛ ✤÷ñ❝
t2 − 6t + 2 + 6 =
16 8
− ⇔ t4 − 6t3 + 8t2 + 8t − 16 = 0
2
t
t
(2)
✣➦t a = 4✱ ♣❤÷ì♥❣ tr➻♥❤ ✭✷✮ trð t❤➔♥❤
✭✸✮
❚❛ ①❡♠ ✭✸✮ ❧➔ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❤❛✐ ✤è✐ ✈î✐ a✳ P❤÷ì♥❣ tr➻♥❤ ♥➔② ❝â
t4 − 6t3 + 8t2 + 8t − 16 = 0 ⇔ a2 − 2ta − t4 + 6t3 − 8t2 = 0.
∆ = t2 + t4 − 6t3 + 8t2 = t2 t2 − 6t + 9 = t2 (t − 3)2 .
❱➟②
(3) ⇔
❉♦ ✤â
a = t + t (t − 3)
a = t − t (t − 3)
t2 − 2t − 4 = 0
⇔
⇔
a = t2 − 2t
a = −t2 + 4t
t=1±
√
5
−t2 + 4t − 4 = 0
t=2
√
√
√
1+ 5
t = 1 + 5✱ t❛ ✤÷ñ❝ t = 1 + 5 = log3 x ⇔ x = 3
✱
❱î✐
✭t❤ä❛ ✤✐➲✉
❦✐➺♥✮✳
√
√
√
1− 5
❱î✐ t = 1 − 5✱ t❛ ✤÷ñ❝ t = 1 − 5 = log3x ⇔ x = 3 ✱ ✭t❤ä❛ ✤✐➲✉
❦✐➺♥✮✳
❱î✐ t = 2✱ t❛ ✤÷ñ❝ 2 = log3x ⇔ x = 32 = 9
√
√
◆❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❧➔✿ x = 31+ 5, x = 31− 5, x = 9✳
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✺✽
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✷✳✺ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ①✉➜t ①ù tø ❤➻♥❤ ❤å❝
❈â ♥❤ú♥❣ ❦➳t q✉↔ ❝õ❛ ❤➻♥❤ ❤å❝ ❝❤♦ ♣❤➨♣ t❛ ①➙② ❞ü♥❣ ♥➯♥ ❝→❝ ❜➔✐
t♦→♥ ✈➲ ♣❤÷ì♥❣ tr➻♥❤✳ ❱➔ ❦❤✐ ❣✐↔✐ ❝→❝ ❜➔✐ t♦→♥ ✤â✱ ♥➳✉ ❜✐➳t ❦❤❛✐ t❤→❝
❝→❝ ✤➦❝ tr÷♥❣ ❤➻♥❤ ❤å❝ ❝õ❛ ♥â s➩ ❝❤♦ t❛ ❝→❝❤ ❣✐↔✐ ❤❛② ❤ì♥ ♥❤ú♥❣ ❝→❝❤
❣✐↔✐ ❦❤→❝✳ ✣➙② ❧➔ ♠ët ❤÷î♥❣ s→♥❣ t→❝ ✤➲ t♦→♥ r➜t t❤ó ✈à✱ ❝❤♦ t❤➜② ♠è✐
❧✐➯♥ ❤➺ ❣✐ú❛ ❤➻♥❤ ❤å❝ ✈➔ ✤↕✐ sè✳
❱➼ ❞ö ✶✳ ❚❛ s➩ ①➙② ❞ü♥❣ ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❞ü❛ tr➯♥ ❦➳t q✉↔✿
→
−
→
−
→
−
−
→
−
−
a , b = 1.
a. b = |→
a | . b ⇔ cos →
❳➨t
→
−
→
−
−
−
⇔ →
a , b = 0 ⇔ ∃k > 0 : →
a = k b.
√
√
a = (x; 1) , b = 2 4x − 3; x + 1 ✳ ❑❤✐ ✤â
√
√
a.b = 2x 4x − 3 + x + 1, |a| . b = (x2 + 1) (17x − 11)
→
−
−
−
❱➟② →
a . b = |→
a |.
→
−
b
♥❣❤➽❛ ❧➔
√
√
2x 4x − 3 + x + 1 =
(x2 + 1) (17x − 11).
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥√✶✳ ●✐↔✐ ♣❤÷ì♥❣
tr➻♥❤
√
2x 4x − 3 +
x+1=
(x2 + 1) (17x − 11)✳
●✐↔✐ ✳ ✣✐➲✉ ❦✐➺♥ x ≥ 43 ✳ ✣➦t a = (x; 1) , b =
✤÷ñ❝
√
√
a.b = 2x 4x − 3 + x + 1, |a| . b =
✭✶✮
√
√
2 4x − 3; x + 1
t❛
(x2 + 1) (17x − 11)
→
−
→
−
→
−
−
−
−
❱➟② ✭✶✮ ❝â ♥❣❤➽❛ ❧➔ →
a. b = |→
a |. b ⇔ →
a = k. b , k ≥ 0✳ ❉♦ ✤â x ❧➔
♥❣❤✐➺♠ ❝õ❛ ✭✶✮ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐ tç♥ t↕✐ k ≥ 0 s❛♦ ❝❤♦
√
√
√
x = 2k 4x − 3
x
1
√
√
⇔
=
⇔
x
x
+
1
=
2
4x − 3
√
2
4x
−
3
x
+
1
1=k x+1
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✺✾
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
⇔
⇔
x≥0
x2 (x + 1) = 4 (4x − 3)
⇔
x≥0
x3 + x2 − 16x + 12 = 0
x≥0
x≥0
⇔
x=3
(x − 3) x2 + 4x − 4 = 0
x = −2 ± 2√2
⇔
x=3
√
x = −2 + 2 2
✭t❤ä❛ ♠➣♥ ✤✐➲✉ ❦✐➺♥✮
P❤÷ì♥❣ tr➻♥❤ ❝â t➟♣ ♥❣❤✐➺♠ ❧➔ S =
√
3; −2 + 2 2
✳
❱➼ ❞ö ✷✳ ❚r♦♥❣ ♠➦t ♣❤➥♥❣ tå❛ ✤ë ❖①②✱ ❝❤♦ ✤✐➸♠ A(a1; a2) ♥➡♠ ♣❤➼❛
❞÷î✐ trö❝ ❖①✱ ✤✐➸♠ B(b1; b2) ♥➡♠ ♣❤➼❛ tr➯♥ trö❝ ❖①✳ ▼ët ✤✐➸♠ M (x; 0)
♥➡♠ tr➯♥ ❖① ✭❤➻♥❤ ✈➩✮✳
❑❤✐ ✤â t❛ ❝â✿
−−→
−−→
−→
M A + M B ≥ AB
❙✉② r❛ t❛ ❝â ✿
(x − a1 )2 + a22 +
(x − b1 )2 + b22 ≥
(a1 − b1 )2 + (a2 − b2 )2
−→
−−→
✣➥♥❣ t❤ù❝ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐✿ −
M A = k.M B ✭❆✱ ▼✱ ❇ t❤➥♥❣ ❤➔♥❣✮✳
❈❤å♥ A(0; −3)✱ B(1, 2)✳ ❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ √x2 + 9+√x2 − 2x + 5 = √26
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✻✵
(∗)
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
●✐↔✐✳ ❚➟♣ ①→❝ ✤à♥❤
D = x ∈ R|x2 − 2x + 5 = (x − 1)2 + 4 ≥ 0 = R
P❤÷ì♥❣ tr➻♥❤ (∗) t÷ì♥❣ ✤÷ì♥❣ ✈î✐
(x − 0)2 + (0 − 3)2 + (x − 1)2 + (0 − 2)2 =
(0 − 1)2 + (−3 − 2)2
❑❤✐ ✤â ❝❤å♥✿ A(0; −3)✱ B(1, 2)✱ M (x; 0)✳ ❙✉② r❛ t❛ ❝â✿
−−→
M A = (x − 0; 0 − 3) = (x; −3)
−−→
M A = (x − 1; 0 − 2) = (x − 1; −2)
−→
AB = (−1; −5)
▼➔ t❛ ❝â
−−→
−−→
−→
M A + M B ≥ AB
✣➥♥❣ t❤ù❝ ①↔② r❛
−−→
−−→
⇔ M A = k.M B ⇔
❱➟② ♣❤÷ì♥❣ tr➻♥❤ (∗) ❝â
x
3
3
=
⇔ −2x = 3x − 3 ⇔ x = .
x − 1 −2
5
♥❣❤✐➺♠ ❞✉② ♥❤➜t x = 35 .
❱➼ ❞ö ✸✳ ❚r♦♥❣ ♠➦t ♣❤➥♥❣ tå❛ ✤ë ❖①②✱ ❝❤♦ ✤✐➸♠ A(a1; a2) ♥➡♠ ♣❤➼❛
❜➯♥ tr→✐ trö❝ ❖②✱ ✤✐➸♠ B(b1; b2) ♥➡♠ ❜➯♥ ♣❤↔✐ trö❝ ❖②✳ ▼ët ✤✐➸♠
M (0; y) ♥➡♠ tr➯♥ trö❝ ❖② ✭❤➻♥❤ ✈➩✮✳
❑❤✐ ✤â t❛ ❝â✿
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
−→
−−→
−−→
AB − M B ≤ M A
✻✶
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❙✉② r❛ t❛ ❝â
(a1 − b1 )2 + (a2 − b2 )2 −
b21 + (b2 − y)2 ≤ a21 + (a2 − y)2
−−→
−−→
M A = k.M B ✭❆✱ ▼✱ ❇ t❤➥♥❣ ❤➔♥❣✮✳
✣➥♥❣ t❤ù❝ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐✿
❈❤å♥ A (2; 0)✱ B (1; 1)✳ ❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
●✐↔✐✳ ❚➟♣ ①→❝ ✤à♥❤ R✳
y 2 − 2y + 2+
4 + y2 =
√
10
(∗)
P❤÷ì♥❣ tr➻♥❤ (∗) t÷ì♥❣ ✤÷ì♥❣ ✈î✐✿
√
y 2 − 2y + 2 + 4 + y 2 = 10
√
⇔ 10 − y 2 − 2y + 2 = 4 + y 2
⇔
(1 + 2)2 + (1 − 0)2 −
=
(−2 − 0)2 + (0 − y)2
(1 − 0)2 + (1 − y)2
❑❤✐ ✤â ❝❤å♥✿ A (−2; 0)✱ B (1; 1)✱ M (0; y)✳ ❙✉② r❛ t❛ ❝â✿
−−→
M A = (−2 − 0; 0 − y)
−−→
M B = (1 − 0; 1 − y)
−→
AB = (3; 1)
→
−−→
−−→
−→
−−→
−−→
❙✉② r❛ (1) ⇔ −
AB − M B = M A ▼➔ t❛ ❝â ✿ AB − M B ≤ M A
❉➜✉ ✤➥♥❣ t❤ù❝ ①↔② r❛✿
❱➟②
−−→
−−→
−2
−y
2
⇔ M A = k.M B ⇔
=
⇔ 3y = 2 ⇔ y = .
1
1−y
3
♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ ❞✉② ♥❤➜t y = 23 ✳
✷✳✻ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ sû ❞ö♥❣ ❜➜t
✤➥♥❣ t❤ù❝
❱➼ ❞ö ✶✳ ❚❛ s➩ sû ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❏❡♥✲s❡♥ ✤➸ s→♥❣ t↕♦ r❛ ♠ët
sè ♣❤÷ì♥❣ tr➻♥❤✳
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✻✷
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❇➜t ✤➥♥❣ t❤ù❝ ❏❡♥✲s❡♥✿
❈❤♦ f (x) ❧➔ ❤➔♠ ❧ç✐ tr➯♥ D✳ ●✐↔ sû✿
x1, x2 , ..., xn ∈ D; αi > 0; i = 1, n
s❛♦ ❝❤♦
n
αi = 1.
i=1
❚❛ ❝â✿
n
f
≤
αi xi
❚r÷í♥❣ ❤ñ♣ ✤➦❝ ❜✐➺t✿
f
n
x1 + x2 + ... + xn
n
i=1
i=1
f (x)
≤
αi f (xi ) .
❧➔ ❤➔♠ ❧ç✐ tr➯♥ D✳ ❚❛ ❝â✿
f (x1 ) + f (x2 ) + ... + f (xn )
n
✈î✐ n ∈ Z∗
❉➜✉ ❜➜t ✤➥♥❣ t❤ù❝ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐ x1 = x2 = ... = xn. ❚r÷í♥❣
❤ñ♣ ❤➔♠ ❧ã♠ t÷ì♥❣ tü ♥❤÷ tr➯♥✳
❳➨t ❤➔♠ sè f (x) = x3 tr➯♥ R+✳ ❚❛ ❝â
f (x) = 3x2
f (x) = 6x > 0, ∀x ∈ R+
❙✉② r❛ f (x) = x3 ❧➔ ❤➔♠ ❧ç✐ tr➯♥ R+
▼➦t ❦❤→❝ →♣ ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❏❡♥✲s❡♥ ✈î✐ ❤❛✐ sè a✱ b ∈ R+ t❛ ❝â
f
a+b
2
f (a) + f (b)
≤
⇔
2
a+b
2
3
a3 + b3
≤
.
2
⇔ (a + b)3 ≤ 4 a3 + b3 .
✣➥♥❣ t❤ù❝ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐ a = b
❈❤♦ a = x, b = 3✱ t❤❛② ✈➔♦ ❜✐➸✉ t❤ù❝ ❝✉è✐ ❝ò♥❣ t❛ ❝â
(x + 3)3 ≤ 4 x3 + 33
⇔ x3 + 9x2 + 27x + 27 ≤ 4x3 + 108
⇔ −3x3 + 9x2 + 27x − 81 ≤ 0, ∀x > 0
❚ø ✤➙② t❛ ❝â ❜➔✐ t♦→♥ s❛✉
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✻✸
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❇➔✐ t♦→♥ ✶✳❚➻♠ ♥❣❤✐➺♠ ♥❣✉②➯♥ ❞÷ì♥❣ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ s❛✉
−3x3 + 9x2 + 27x − 81 = 0 (1)
●✐↔✐✳ ❚❛ ❝â
(1) ⇔ x3 + 9x2 + 27x + 27 = 4x3 + 108
⇔ (x + 3)3 = 4 x3 + 33
3
x3 + 3 3
x+3
⇔
≤
2
2
▼➦t ❦❤→❝✱ →♣ ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❏❡♥✲s❡♥ ❝❤♦ ❤➔♠ sè f (x) = x3 ❧➔
❤➔♠ ❧ç✐ tr➯♥ R+ ✭ ✈➻ f (x) = 6x > 0, ∀x ∈ R+✮ ✈î✐ ❤❛✐ sè x ✈➔ 3 t❛ ❝â
f
x+3
2
≤
f (x) + f (3)
2
3
x3 + 33
(2)
2
❱➟② (1) ⇔ ❞➜✉ ✧❂✧ ð ✭✷✮ ①↔② r❛ ⇔ x = 3. ❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠
x+3
2
≤
❞✉② ♥❤➜t x = 3✳
❱➼ ❞ö ✷✳ ❳➨t ❤➔♠ sè f (x) = x5 tr➯♥ R+✳ ❚❛ ❝â
f (x) = 5x4
f (x) = 20x3 > 0, ∀x ∈ R+
❙✉② r❛ f (x) = x5 ❧➔ ❤➔♠ ❧ç✐ tr➯♥ R+
▼➦t ❦❤→❝✱ →♣ ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❏❡♥✲s❡♥ ✈î✐ ❤❛✐ sè a✱ b ∈ R+ t❛ ❝â
a+b
f (a) + f (b)
f
≤
⇔
2
2
⇔ (a + b)5 ≤ 24 a5 + b5 .
a+b
2
5
a5 + b5
≤
2
✣➥♥❣ t❤ù❝ ①↔② r❛ ⇔ a = b
❈❤♦ a = x, b = 3✱ t❤❛② ✈➔♦ ❜✐➸✉ t❤ù❝ ❝✉è✐ ❝ò♥❣ t❛ ❝â
(x + 3)5 ≤ 16 x5 + 35
⇔ 15x5 − 15x4 − 90x3 − 270x2 − 405x + 3645 ≥ 0.
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✻✹
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❚ø ✤➙② t❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✷✳ ❚➻♠ ♥❣❤✐➺♠ ♥❣✉②➯♥ ❞÷ì♥❣ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ s❛✉
15x5 − 15x4 − 90x3 − 270x2 − 405x + 3645 = 0
❍÷î♥❣ ❞➝♥✳
15x5 − 15x4 − 90x3 − 270x2 − 405x + 3645 = 0
⇔ x5 + 15x4 + 90x3 + 270x2 + 405x + 243 = 16x5 + 3888
⇔ (x + 3)5 = 24 x5 + 35
5
x5 + 3 5
x+3
⇔
=
2
2
▼➦t ❦❤→❝✱ →♣ ❞ö♥❣ ❜➜t ✤➥♥❣ t❤ù❝ ❏❡♥✲s❡♥ ❝❤♦ ❤➔♠ sè f (x) = x5 ❧➔
❤➔♠ ❧ç✐ tr➯♥ R+ ✈î✐ ❤❛✐ sè x ✈➔ 3 t❛ ❝â
x+3
2
5
x5 + 3 5
≤
2
✣➥♥❣ t❤ù❝ ①↔② r❛ ❦❤✐ x = 3 ❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ ❞✉② ♥❤➜t
x = 3✳
❱➼ ❞ö ✸✳ ❚❛ s➩ ①➨t ❤❛✐ ❜➜t ✤➥♥❣ t❤ù❝ ❝â ❞➜✉ ❜➡♥❣ ❝ò♥❣ ①↔② r❛ ❦❤✐
x = 2✱
❝❤➥♥❣ ❤↕♥ ✈î✐ x ∈
√
5
− ; +∞
2
✱ t❛ ❝â
32 + 2x + 5
+ 5) ≤
= 7 + x (i)
3 2x + 5 =
2
x3 − 12x + 16 = (x − 2)2 (x + 4) ≥ 0 (ii)
32 (2x
❱î✐ x ∈ − 25 ; +∞ t❤➻ ❞➜✉ ❜➡♥❣ ð ✭✐✮ ✈➔ ✭✐✐✮ ❝ò♥❣ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾
❦❤✐ x = 3✳ ❚ø ✭✐✮✱ ✭✐✐✮ ✈➔ (x3 − 12x + 16) + (7 + x) = x3 − 11x + 23✱
t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
x3 − 11x + 23 − 3 2x + 5 = 0
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✻✺
(1)
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x ≥ − 52 ✳ ❳➨t ❤❛✐ ❤➔♠ sè tr➯♥
5
− ; +∞
2
❧➔
√
f (x) = x3 − 11x + 23, g (x) = 3 2x + 5
❚❤❡♦ ❜➜t ✤➥♥❣ t❤ù❝ ❈❛✉❝❤②✱ t❛ ❝â
g (x) =
32 (2x + 5) ≤
32 + 2x + 5
= 7 + x. (2)
2
❉➜✉ ❜➡♥❣ ð ✭✷✮ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐ x = 2✳ ▼➦t ❦❤→❝
f (x)−(x+7) = x3 −12x+16 = (x − 2)2 (x + 4) ≥ 0, ∀x ≥ −
❉➜✉ ❜➡♥❣ ð ✭✸✮ ①↔② r❛ ❦❤✐ ✈➔ ❝❤➾ ❦❤✐ x = 2✳ ❚❛ ❝â
(1) ⇔ f (x) = g (x) .
5
2
✭✸✮
(4)
❱➟② ✭✹✮ ❝â ♥❣❤➽❛ ❧➔ ❞➜✉ ❜➡♥❣ ð ✭✷✮ ✈➔ ✭✸✮ ✤ç♥❣ t❤í✐ ①↔② r❛✱ ❤❛② x = 2
✭t❤ä❛ ♠➣♥ ✤✐➲✉ ❦✐➺♥✮✳ P❤÷ì♥❣ tr➻♥❤ ✭✶✮ ❝â ♥❣❤✐➺♠ ❞✉② ♥❤➜t x = 2✳
❱➼ ❞ö ✹✳ ❳➨t ♣❤÷ì♥❣ tr➻♥❤
3
x3 + ax2 + bx + c = d x2 − 2x − 1 + x + e(d > 0).
√
P❤÷ì♥❣ tr➻♥❤ ♥➔② ❝â ✤✐➲✉ ❦✐➺♥ x2 − 2x − 1 ≥ 0✳ ❉♦ d x2 − 2x − 1 ≥ 0
♥➯♥
√
3
x3 + ax2 + bx + c ≥ x + e
⇔ x3 + ax2 + bx + c ≥ x + 3ex2 + 3e2 x + e3
⇔ (a − 3e) x2 + (b − 3e2 )x + c − e3 ≥ 0.
❚❛ ❝➛♥ ❝❤å♥ ❛✱❜✱❝✱❡ s❛♦ ❝❤♦
a − 3e = −1
b − 3e2 = 2
⇒
c − e3 = 1
a = 3e − 1
b = 3e2 + 2
c = e3 + 1.
❈❤➥♥❣ ❤↕♥ ❝❤å♥ e = 2✱ d > 0 tò② þ✱ ❝❤➥♥❣ ❤↕♥ d = 2✳ ❑❤✐ ✤â a = 5✱
b = 14✱ c = 9✳ ❚❛ ✤÷ñ❝ ♣❤÷ì♥❣ tr➻♥❤ ❣✐↔✐ ❜➡♥❣ ♣❤÷ì♥❣ ♣❤→♣ ✤→♥❤ ❣✐→
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✻✻
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
♥❤÷ s❛✉✳
❇➔✐ t♦→♥ ✹✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
3
x3 + 5x2 + 14x + 9 − x = 2
x2 − 2x − 1 + 1
(1)
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x2 − 2x − 1 ≥ 0✳ ❚❛ ❝â
3
❉♦
√
x3 + 5x2 + 14x + 9 = 2 x2 − 2x − 1 + 2 + x
(2)
x2 − 2x − 1 ≥ 0✱ ♥➯♥ tø ✭✷✮ t❛ ❝â
√
3
x3 + 5x2 + 14x + 9 ≥ 2 + x
⇔ x3 + 5x2 + 14x + 9 ≥ 8 + 12x + 6x2 + x3
⇔ x2 − 2x − 1 ≤ 0.
❚ø ✤➙② ❦➳t ❤ñ♣ ✈î✐ ✤✐➲✉ ❦✐➺♥ x2 − 2x − 1 ≥ 0, s✉② r❛ ✭✶✮ t÷ì♥❣ ✤÷ì♥❣
✈î✐
√
x2 − 2x − 1 = 0
2
⇔ x − 2x − 1 = 0 ⇔ x = 1 ± 2.
√
3
x3 + 5x2 + 14x + 9 = 2 + x
❱➼ ❞ö ✺✳ ❚ø ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ♥➔♦ ✤â✱ ❝❤➥♥❣ ❤↕♥
4t3 + 3t = 2✳
✣➦t t = 2x✱ t❛ ✤÷ñ❝ 32x3 + 6x = 2 ⇔ 16x3 + 3x − 1 = 0✳
❳➨t ♣❤÷ì♥❣ tr➻♥❤
3
✣✐➲✉
αx3 + ax2 + bx + c = d
16x3 + 3x − 1 + x + e, (d > 0).
√
❦✐➺♥ 16x3 + 3x − 1 ≥ 0. ❉♦ 16x3 + 3x − 1 ≥ 0 ♥➯♥
√
3
αx3 + ax2 + bx + c ≥ x + e
⇔ αx3 + ax2 + bx + c ≥ x3 + 3ex2 + 3e2 x + e3
⇔ (α − 1) x2 + (a − 3e)x2 + (b − 3e2 )x + c − e3 ≥ 0.
❚❛ ❝➛♥ ❝❤å♥ a, b,c, e s❛♦ ❝❤♦
α − 1 = −16
α = −15
a − 3e = 0
a = 3e
⇒
2
b − 3e = −3
b = 3e2 − 3
c − e3 = 1
c = e3 + 1
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✻✼
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❈❤➥♥❣ ❤↕♥ ❝❤å♥ e = 1✱ d = 71✱ t❛ ✤÷ñ❝ ♣❤÷ì♥❣ tr➻♥❤
3
−15x3 + 3x2 + 2 = 71 16x3 + 3x − 1 + x + 1
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✺✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
3
−15x3 + 3x2 + 2 = 71 16x3 + 3x − 1 + x + 1
(1)
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ 16x3 + 3x − 1 ≥ 0✳ ❚❛ ❝â
❉♦
(1) ⇔ −15x3 + 3x2 + 2 = 71 16x3 + 3x − 1 + x + 1.
√
71 16x3 + 3x − 1 ≥ 0 ♥➯♥ tø ✭✶✮ s✉② r❛
√
(1) ⇔ −15x3 + 3x2 + 2 ≥ x + 1
⇔ −15x3 + 3x2 + 2 ≥ x3 + 3x2 + 3x + 1
⇔ 16x3 + 3x − 1 ≤ 0.
❚ø ✤➙② ❦➳t ❤ñ♣ ✈î✐ ✤✐➲✉ ❦✐➺♥✱ t❛ ✤÷ñ❝ 16x3 + 3x − 1 = 0. ❉♦ ✤â
(1) ⇔
16x3 + 3x − 1 = 0
⇔ 16x3 + 3x − 1 = 0.
√
3
2
−15x + 3x + 2 = x + 1
✣➦t x = 2t ✱ t❤❛② ✈➔♦ ✭✸✮ t❛ ✤÷ñ❝
4t3 + 3t = 2.
❱➻ ❤➔♠ sè f (t) = 4t3 + 3t ❝â f (t) = 12t2 + 3 > 0✱∀t ∈ R ♥➯♥ f ✤ç♥❣
❜✐➳♥✱ s✉② r❛ ✭✹✮ ❝â ❦❤æ♥❣ q✉→ ♠ët ♥❣❤✐➺♠✳ ❳➨t
1
1
2=
α3 − 3
2
α
❉♦ ✤â✱ ♥➳✉ ✤➦t α =
3
1
1
2=
α3 − 3
2
α
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
⇔ α
2+
√
3 2
5
3
3
− 4α − 1 = 0 ⇔ α = 2 ±
t❤➻ 2 = 12
1
1
=3
α−
2
α
✻✽
α3 −
1
α3
√
5.
✳ ❚❛ ❝â
1
1
+4
α−
2
α
3
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
√
√
❱➟② t = 12 α − α1 = 21 2 + 5 + 2 − 5 ❧➔ ♠ët ♥❣❤✐➺♠ ❝õ❛
✭✹✮ ✈➔ ❝ô♥❣ ❧➔ ♥❣❤✐➺♠ ❞✉② ♥❤➜t ❝õ❛ ✭✹✮✳ P❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â
♥❣❤✐➺♠ ❞✉② ♥❤➜t
3
x=
1
1
α−
4
α
=
1
4
3
3
2+
√
5+
3
√
2−
5 .
✷✳✼ ▼ët sè ❤÷î♥❣ s→♥❣ t↕♦ ❝→❝ ♣❤÷ì♥❣ tr➻♥❤ ✤❛ t❤ù❝ ❜➟❝
❝❛♦
✷✳✼✳✶ ❙û ❞ö♥❣ ♥❤ú♥❣ ♣❤÷ì♥❣ tr➻♥❤ ❝❤å♥ tr÷î❝
❱➼ ❞ö ✶✳ ❚❛ ①➨t ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❞↕♥❣ a(bx2 + c)2 = (dx + e)2✱
❝❤➥♥❣ ❤↕♥ ♣❤÷ì♥❣ tr➻♥❤ 5 2x2 − 1 2 = (x − 3)2 ⇔ 5
x2 − 6x + 9✳ ❚❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
4x4 − 4x2 + 1 =
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ 20x4 − 21x2 + 6x − 4 = 0✳
●✐↔✐✳ ❚➟♣ ①→❝ ✤à♥❤ R✳ P❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t÷ì♥❣ ✤÷ì♥❣
2
5 4x4 − 4x2 + 1 = x2 − 6x + 9 ⇔ 5 2x2 − 1 = (x − 3)2
√
√
√
√
2 5x2 − x + 3 − 5 = 0
(1)
2 5x2 − 5 = x − 3
⇔
⇔
√ 2 √
√ 2
√
2 5x − 5 = 3 − x
2 5x + x − 3 − 5 = 0
(2)
P❤÷ì♥❣ tr➻♥❤ ✭✶✮ ✈æ ♥❣❤✐➺♠✱ ♣❤÷ì♥❣ tr➻♥❤ ✭✷✮ ❝â ♥❣❤✐➺♠ ❧➔
x=
−1 +
√
41 + 24 5
−1 −
,x =
2
❱➼ ❞ö ✷✳ ❙û ❞ö♥❣ ♣❤÷ì♥❣ tr➻♥❤
√
41 + 24 5
.
2
a6 − b 6
= 0✱ ✈î✐ a = x+1✱ b = x−1✱
a−b
t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
(x + 1)5 + (x + 1)4 (x − 1) + (x + 1)3 (x − 1)2
+(x + 1)2 (x − 1)3 + (x + 1) (x − 1)4 + (x − 1)5 = 0 (1)
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✻✾
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
●✐↔✐✳ ✣➦t a = x + 1✱ b = x − 1✳ ❑❤✐ ✤â✱ t❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝
a5 + a4 b + a3 b2 + a2 b3 + ab4 + b5 = 0.
(2)
❚❛ ❝â ❤➡♥❣ ✤➥♥❣ t❤ù❝
a6 − b6 = (a − b) (a5 + a4 b + a3 b2 + a2 b3 + ab4 + b5 ) = 0.
6
6
b
❚ø ✭✷✮ t❛ ❝â aa −
= 0 ⇔ a6 = b6 ⇔ a = −b ✭❞♦ a = b✮✳
−b
◆❤÷ ✈➟② (1) ⇔ x + 1 = 1 − x ⇔ x = 0. P❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ ❞✉②
♥❤➜t x = 0✳
✷✳✼✳✷ ❙û ❞ö♥❣ ❝æ♥❣ t❤ù❝ ❧÷ñ♥❣ ❣✐→❝
❚r♦♥❣ ♠ö❝ ♥➔② t❛ s➩ ❞ò♥❣ ♠ët sè ❝æ♥❣ t❤ù❝ ❧÷ñ♥❣ ❣✐→❝ ✤➸ s→♥❣ t→❝ r❛
❝→❝ ♣❤÷ì♥❣ tr➻♥❤ ✤❛ t❤ù❝ ❜➟❝ ❝❛♦✳ ❱✐➺❝ ❣✐↔✐ ❝→❝ ♣❤÷ì♥❣ tr➻♥❤ ✤❛ t❤ù❝
❜➟❝ ❝❛♦ ❧➔ r➜t ♣❤ù❝ t↕♣✱ tr♦♥❣ ♥❤✐➲✉ tr÷í♥❣ ❤ñ♣ ❧➔ ❦❤æ♥❣ t❤➸✳ ❚✉②
♥❤✐➯♥✱ sû ❞ö♥❣ t➼♥❤ ❝❤➜t ♣❤÷ì♥❣ tr➻♥❤ ✤❛ t❤ù❝ ❜➟❝ n (n = 1, 2, ...) ❝â
❦❤æ♥❣ q✉→ n ♥❣❤✐➺♠✱ ✈➔ ♠ët sè ✤à♥❤ ❤÷î♥❣ tr♦♥❣ q✉→ tr➻♥❤ s→♥❣ t→❝
✤➲ t♦→♥✱ t❛ ❝â ✤÷ñ❝ ❧í✐ ❣✐↔✐ r➜t ♥❣➢♥ ❣å♥ ✈➔ ➜♥ t÷ñ♥❣ ❝❤♦ ❝→❝ ♣❤÷ì♥❣
tr➻♥❤ ❞↕♥❣ ♥➔②✳
❱➼ ❞ö ✶✳ ❚ø ❝æ♥❣ t❤ù❝ cos 6α = 32cos6α − 48cos4α + 18cos2α − 1✱
❧➜② cos α = x✱ t❛ ✤÷ñ❝
❈❤å♥ 6α = π3 ✱ t❛
cos 6α = 32x6 − 48x4 + 18x2 − 1.
✤÷ñ❝ 32x6 − 48x4 + 18x2 − 1 = 21 . ❚❛
❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ✭✣➲ ♥❣❤à ❖❧②♠♣✐❝ ✸✵✴✵✹✴✷✵✵✾✮✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
64x6 − 96x4 + 36x2 − 3 = 0.
●✐↔✐✳ ❚❛ ❝â
cos 6α = 2cos2 3α − 1 = 2 4cos3 α − 3 cos α
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✼✵
2
−1
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
= 32cos6 α − 48cos4 α + 18cos2 α − 1
(1)
P❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t÷ì♥❣ ✤÷ì♥❣
32x6 − 48x4 + 18x2 − 1 =
1
π
⇔ 32x6 − 48x4 + 18x2 − 1 = cos .
2
3
(2)
k2π
π
❚ø ✭✶✮ s✉② r❛ ✭✷✮❝â ✻ ♥❣❤✐➺♠ ❧➔ x = cos 3.6
+
, k = 0, 1, 2, 3, 4, 5.
6
▲÷✉ þ✳ ❱✐➺❝ ♥❤î ❝æ♥❣ t❤ù❝ ❜✐➸✉ ❞✐➵♥ cos nα t❤❡♦ cos α✱ sin nα t❤❡♦
sin α s➩ ❣✐ó♣ t❛ ❣✐↔✐ ✤÷ñ❝ ♥❤ú♥❣ ♣❤÷ì♥❣ tr➻♥❤ ❞↕♥❣ ♥➔②✳
❱➼ ❞ö ✷✳ ❚ø cos 5α = 16cos5α − 20cos3α + 5 cos α✱ ✤➦t cos α = 2√x 3
t❛ ✤÷ñ❝
16x5
20x3
5x
x5
5x3
5x
x5 − 15x3 + 45x
√ − √ + √ = √ − √ + √ =
√
.
288 3 24 3 2 3 18 3 6 3 2 3
18 3
√
π
3 x5 − 15x3 + 45x
√
⇔ x5 −15x3 +45x−27 = 0✳
❈❤å♥ 5α = 6 ✤÷ñ❝ 2 =
18 3
cos 5α =
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ x√5 − 15x3 + 45x − 27 = 0.
●✐↔✐✳ ❚➟♣ ①→❝ ✤à♥❤ R✳ ✣➦t x = 2 3t✱ t❤❛② ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦
t❛ ✤÷ñ❝
√
√
√
288 3t5 − 360 3t3 + 90 3t − 27 = 0
√
⇔ 2 16t5 − 20t3 + 5t = 3
π
(1)
⇔ 16t5 − 20t3 + 5t = cos .
6
▼➦t ❦❤→❝ t❛ ❝â
cos 5α + cos α = 2 cos 3α cos 2α
⇔ cos 5α = 2 4cos3 α − 3 cos α
2cos2 α − 1 − cos α
⇔ cos 5α = 2 8cos5 α − 10cos3 α + 3 cos α − cos α
⇔ cos 5α = 16cos5 α − 20cos3 α + 5 cos α.
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✼✶
(2)
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚ø ✭✷✮ s✉② r❛ ✭✶✮ ❝â ✺ ♥❣❤✐➺♠ ❧➔ t = cos
P❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ✺ ♥❣❤✐➺♠ ❧➔
π
k2π
+
6.5
5
√
x = 2 3 cos
k2π
π
+
30
5
, k = 0, 1, 2, 3, 4.
, k = 0, 1, 2, 3, 4.
▲÷✉ þ✳ ❚r♦♥❣ ❧í✐ ❣✐↔✐ tr➯♥✱ ♣❤➨♣ ✤➦t x = 2√3t t➻♠ r❛ ♥❤÷ s❛✉✿ ❉♦
❝æ♥❣ t❤ù❝ cos 5α = 16cos5α − 20cos3α + 5 cos α✱ ♥➯♥ ✤➦t x = at✱ ✈î✐ ❛
s➩ t➻♠ s❛✉✳ ❚❤❛② x = at ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦✱ t❛ ✤÷ñ❝
a5 t5 − 15a3 t3 + 45at − 27 = 0.
❚❛ t➻♠ a t❤ä❛ ♠➣♥ ✤✐➲✉ ❦✐➺♥
❱➟② t❛
√
a5
15a3
45a
a4
3a3
=
=
⇒
=
= 9 ⇒ a = ±2 3.
16
20
5
16
4
√
❝â ♣❤➨♣ ✤➦t x = 2 3t✳
❱➼ ❞ö ✸✳ ❚ø sin5α = 16sin5α − 20sin3α + 5 sin α✱ ❧➜② sin α = 2x t❛
✤÷ñ❝ sin5α = 512x5 − 160x3 + 10x✳ ❈❤å♥ 5α = π3 ✱ t❛ ❝â
√
√
3
= 512x5 − 160x3 + 10x ⇔ 1024x5 − 320x3 + 20x − 3 = 0.
2
❚❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ tâ❛♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ 1024x5 − 320x3 + 20x − √3 = 0✳
●✐↔✐✳ ✣➦t x = 2t ✱ t❤❛② ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t❛ ✤÷ñ❝
32t5 − 40t3 + 10 =
❚❛ ❝â
√
π
3 ⇔ 16t5 − 20t3 + 5t = sin .
3
sin 5α + sin α = 2 sin 3α cos 2α
⇔ sin 5α = 2 3 sin α − 4sin3 α
1 − 2sin2 α − sin α
⇔ sin 5α = 2 8sin5 α − 10sin3 α + 3 sin α − sin α
⇔ sin 5α = 16sin5 α − 20sin3 α + 5 sin α.
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✼✷
(2)
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚ø ✭✷✮ s✉② r❛ ✭✶✮ ❝â ✺ ♥❣❤✐➺♠ ❧➔ t = sin
P❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ✺ ♥❣❤✐➺♠ ❧➔
π
k2π
+
3.5
5
x=
1
k2π
π
sin
+
2
15
5
, k = 0, 1, 2, 3, 4.
, k = 0, 1, 2, 3, 4
✷✳✼✳✸ ❙û ❞ö♥❣ ♥❤à t❤ù❝ ◆✐✉✲tì♥
❈æ♥❣ t❤ù❝ ♥❤à t❤ù❝ ◆✐✉✲tì♥ ✭❣å✐ t➢t ❧➔ ♥❤à t❤ù❝ ◆✐✉✲tì♥ ❧➔✮✿
(a + b)n = Cn0 an + Cn1 an−1 b + ... + Cnk an−k bk + ... + Cnn bn
n
Cnk an−k bk ✭q✉②
=
÷î❝ a0 = b0 = 1)
k=0
❙❛✉ ✤➙②✱ t❛ ❝ò♥❣ ♥❣❤✐➯♥ ❝ù✉ ❝→❝ ✈➼ ❞ö s→♥❣ t→❝ ❜➔✐ t♦→♥ tø ♥❤à t❤ù❝
◆✐✉✲tì♥✳
❱➼ ❞ö ✶✳ ❚❛ ❝â
√ 3
√
10
1
3
10 5
0
2
4 2
( x) + C10
x.
+ C10
x + C10
x + ... + C10
x) = C10
x + C10
√ 10
√
√
3
1
3
10 5
0
2
4 2
(1 − x) = C10
− C10
x + C10
x − C10
( x) + C10
x − ... − C10
x.
(1 +
√
❈ë♥❣ ❧↕✐ t❛ ✤÷ñ❝
1+
√
x
10
+ 1−
√
x
10
0
2
4 2
10 5
= 2 C10
+ C10
x + C10
x + ... + C10
x
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
●✐↔✐✳ ❳➨t
❙✉② r❛
√
1 + i −x
x5 + 45x4 + 210x3 + 210x2 + 45x + 1 = 0.
√
2k
2k
2k
2k k
= C20
(−1)k (−x)k = C20
x ✳
x < 0✳ ❚❛ ❝â C20
i −x
10
√
+ 1 − i −x
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
10
0
2
4 2
10 5
= 2 C10
+ C10
x + C10
x + ... + C10
x .
✼✸
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❉♦ ✤â
0
2
4 2
10 5
(1) ⇔ 2 C10
+ C10
x + C10
x + ... + C10
x =0
√
10
√
√
1 + i −x
10
10
√
= −1.
⇔ 1 + i −x + 1 − i −x = 0 ⇔
1 − i −x
√
1 + i −x
◆❣❤➽❛ ❧➔ 1 − i√−x ❧➔ ♠ët ❝➠♥ ❜➟❝ ♠÷í✐ ❝õ❛ −1 = cos π + i sin π. ❉♦
✤â
√
1 + i −x
π + k2π
π + k2π
√
+ i sin
, k = 0, 1, 2, ..., 9.
= cos
10
10
1 − i −x
❚❛ ❝â
√
1 − i −x
(2)
π + k2π
π + k2π
+ i sin
10
10
π + k2π
π + k2π √
+ −x sin
= cos
10
10
√
π + k2π
π + k2π
− −x cos
i.
+ sin
10
10
cos
❚ø ✭✷✮ t❛ ♥❤➙♥ ❝❤➨♦✱ s❛✉ ✤â ✤ç♥❣ ♥❤➜t ♣❤➛♥ t❤ü❝ ✈î✐ ♣❤➛♥ t❤ü❝✱ ♣❤➛♥
↔♦ ✈î✐ ♣❤➛♥ ↔♦ ✤÷ñ❝ ❦➳t q✉↔
√
−x = tan
√
π + k2π
, k = 0, 1, 2, ..., 9.
10
(3)
❉♦ −x > 0 ♥➯♥ tr♦♥❣ ✭✸✮ t❛ ❝❤➾ ❝❤♦ k ❝❤↕② tø ✵ ✤➳♥ ✹✳ ❱➟② ♣❤÷ì♥❣
tr➻♥❤ ✭✶✮ ❝â ✺ ♥❣❤✐➺♠ ❧➔
x = −tan2
π + k2π
, k = 0, 1, 2, 3, 4.
10
(4)
◆❤÷♥❣ ✭✶✮ ❧➔ ♣❤÷ì♥❣ tr➻♥❤ ✤❛ t❤ù❝ ❜➟❝ ✺ ♥➯♥ ❝â ❦❤æ♥❣ q✉→ ✺ ♥❣❤✐➺♠✳
❱➟② ✭✹✮ ❝❤♦ t❛ t➜t ❝↔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ✭✶✮✳
❱➼ ❞ö ✷✳ ❙û ❞ö♥❣
1+
√
12
t
+ 1−
√
12
t
0
2
4 2
12 6
= 2 C12
+ C12
t + C12
t + ... + C12
t ,
t❛ t❤✉ ✤÷ñ❝ ♣❤÷ì♥❣ tr➻♥❤
t6 + 66t5 + 495t4 + 924t3 + 495t2 66t + 1 = 0.
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✼✹
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
▲↕✐ ❝❤❡ ❞➜✉ ❦➽ ❤ì♥ ❜➡♥❣ ❝→❝❤ ✤➦t t = −2x✱ ❞➝♥ tî✐ ♣❤÷ì♥ tr➻♥❤
64t6 − 2112t5 + 7920t4 − 7392t3 + 1980t2 − 132t + 1 = 0.
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
64t6 − 2112t5 + 7920t4 − 7392t3 + 1980t2 − 132t + 1 = 0.
❍÷î♥❣ ❞➝♥✳ ✣➦t x = pt✱ t❤❛② ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤✱ ❞➝♥ tî✐ ❝➛♥ ♣❤↔✐
❝❤å♥ p = 2✳▼ët ❞➜✉ ❤✐➺✉ ❦❤→❝ ❝ô♥❣ rã r➔♥❣✱ ✤â ❧➔ ❤➺ sè tü ❞♦ ❧➔ ✶✱ sè
❤↕♥❣ ❝❤ù❛ ❧ô② t❤ø❛ ❝❛♦ ♥❤➜t ❧➔ 64x6 = (−2x)6✳
✷✳✽ ▼ët sè ❤÷î♥❣ s→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ♣❤➙♥ t❤ù❝ ❤ú✉ t➾
❚r♦♥❣ ♣❤➛♥ ♥➔② tr➻♥❤ ❜➔② ♣❤÷ì♥❣ ♣❤→♣ ①➙② ❞ü♥❣ ♣❤÷ì♥❣ tr➻♥❤ tø
♥❤ú♥❣ ✤➥♥❣ t❤ù❝ ✤↕✐ sè ❝â ✤✐➲✉ ❦✐➺♥ ✲ ♠ët tr♦♥❣ ♥❤ú♥❣ ♣❤÷ì♥❣ ♣❤→♣
❣✐ó♣ ❝❤ó♥❣ t❛ t↕♦ r❛ ♥❤ú♥❣ ♣❤÷ì♥❣ tr➻♥❤ ♣❤➙♥ t❤ù❝ ❤ú✉ t➾ ❤❛② ✈➔ ❧↕✳
✶✳ ❉ò♥❣ ❤➡♥❣ ✤➥♥❣ t❤ù❝
(a + b + c)3 = a3 + b3 + c3 + 3 (a + b) (b + c) (c + a)
(∗)
❈❤ù♥❣ ♠✐♥❤✳ ❚❛ ❝â✿
(a + b + c)3 − a3 − b3 − c3 = (a + b + c)3 − a3 − b3 + c3
= (b + c) (a + b + c)2 + a (a + b + c) + a2 − (b + c) b2 − bc + c2
= (b + c) (a + b + c)2 + a (a + b + c) + a2 − b2 − bc + c2
= (b + c) 3a2 + 3ab + 3bc + 3ac = 3 (b + c) a2 + ab + bc + ca
= 3 (a + b) (b + c) (c + a)
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✼✺
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❚ø ❤➡♥❣ ✤➥♥❣ t❤ù❝ ✭✯✮ t❤➜② ♥❣❛② r➡♥❣
a+b=0
(a + b + c)3 = a3 +b3 +c3 ⇔ (a + b) (b + c) (c + a) = 0 ⇔
b+c=0
c + a = 0.
❱➼ ❞ö ✶✳ ❈❤å♥ a, b, c✱ ❝❤➥♥❣ ❤↕♥ a = x − 2, b = 2x − 4, c = 7 − 3x✳
❙❛✉ ✤â ❝❤♦ (a + b + c)3 = a3 + b3 + c3 t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ (x − 2)3 +(2x − 4)3 +(7 − 3x)3 = 1.(1)
●✐↔✐✳ ✣➦t a = x − 2, b = 2x − 4, c = 7 − 3x✳ ❚❤❛② ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝
✭✷✮
(a + b + c)3 = a3 + b3 + c3
▼➦t ❦❤→❝ t❛ ❧✉æ♥ ❝â ✤➥♥❣ t❤ù❝
(a + b + c)3 = a3 + b3 + c3 + 3 (a + b) (b + c) (c + a) .
(3)
❚ø ✭✷✮ ✈➔ ✭✸✮ t❛ ❝â
(a + b) (b + c) (c + a) = 0
❉♦ ✤â
(1) ⇔ (x − 2)3 + (2x − 4)3 + (7 − 3x)3
= [(x − 2) + (2x − 4) + (7 − 3x)]3
❱➟②
x=2
⇔ (3x − 6) (3 − x) (5 − 2x) = 0 ⇔ x = 3
5
x=
2
♣❤÷ì♥❣ tr➻♥❤ ✭✶✮ ❝â t➟♣ ♥❣❤✐➺♠ S = 2, 3, 25 .
❱➼ ❞ö ✷✳ ❈❤å♥ a = x2 − 4x + 1, b = 8x − x2 + 4, c = x − 5✳ ❙❛✉ ✤â
❝❤♦ (a + b + c)3 = a3 + b3 + c3 t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
(x2 − 4x + 1)3 + (8x − x2 + 4)3 +(x − 5)3 = 125x3
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✼✻
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✷✳ ❉ò♥❣ ♠➺♥❤ ✤➲
1
1 1 1
+ + =
⇔ (a + b) (b + c) (c + a) = 0
a b c
a+b+c
❈❤ù♥❣ ♠✐♥❤✳ ❚❛ ❝â
(a + b) (b + c) (c + a) = 0 ⇔ (a + b) bc + ba + ac + c2 = 0
⇔ 2abc + a2 b + a2 c + b2 c + b2 a + c2 a + c2 b = 0.
(1)
▼➦t ❦❤→❝
1 1 1
1
ab + bc + ca
1
+ + =
⇔
=
a b c
a+b+c
abc
a+b+c
⇔ (a + b + c) (ab + bc + ca) = abc
2
2
2
2
2
(2)
2
⇔ a b + a c + abc + b c + b a + abc + c a + c b = 0
⇔ 2abc + a2 b + a2 c + b2 c + b2 a + c2 a + c2 b = 0.
❚ø ✭✶✮ ✈➔ ✭✷✮ s✉② r❛ ♠➺♥❤ ✤➲ tr➯♥ ✤ó♥❣✳
❱➼ ❞ö ✶✳ ❚ø ♠➺♥❤ ✤➲ tr➯♥✱ ❧➜② a = x − 8, b = 2x + 7, c = 5x + 8✱ t❛
✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ x −1 8 + 2x 1+ 7 + 5x 1+ 8 = 8x 1+ 7 (1)
−8
−7
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ ✿ x = 8, x = −7
,x =
,x =
✳ ❚❛ ❝❤ù♥❣ ♠✐♥❤
2
5
8
✤÷ñ❝
1 1 1
1
+ + =
⇔ (a + b) (b + c) (c + a) = 0
a b c
a+b+c
(2)
❉♦ ✭✷✮ ♥➯♥ t❛ ❝â
(1) ⇔ (x − 8 + 2x + 7) (x − 8 + 5x + 8) (2x + 7 + 5x + 8) = 0
1
x=
3
⇔x=0
15
x=− .
7
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✼✼
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
P❤÷ì♥❣ tr➻♥❤ ❝â t➟♣ ♥❣❤✐➺♠ ❧➔ S =
1
15
; 0; −
3
7
▲÷✉ þ✳ ✣è✐ ✈î✐ ❜➔✐ t♦→♥✱ ♥➳✉ ❦❤æ♥❣ ❜✐➳t sû ❞ö♥❣ ♠➺♥❤ ✤➲ ✭✷✮ t❤➻ s➩
r➜t ❦❤â ❦❤➠♥ ✤➸ t➻♠ r❛ ❧í✐ ❣✐↔✐✳
✸✳ ❉ò♥❣ ♠➺♥❤ ✤➲ ✿
✧◆➳✉ xyz = 1 ✈➔ x + y + z = x1 + y1 + 1z t❤➻ (x − 1) (y − 1) (z − 1) = 0✧
❈❤ù♥❣ ♠✐♥❤✳ ❚ø ❣✐↔ t❤✐➳t t❛ ❝â xyz = 1 ✈➔ x + y + z = xy + yz + zx✳
❉♦ ✤â
(x − 1) (y − 1) (z − 1) = (x − 1) (yz − y − z + 1)
= xyz − xy − xz + x − yz + y + z − 1 = x + y + z − (xy + yz + zx) = 0.
❱➼ ❞ö ✶✳ ❈❤å♥ a, b, c s❛♦ ❝❤♦ abc = 1✱ ❝❤➥♥❣ ❤↕♥
a = 2x − 1, b = 5x − 3, c =
❑❤✐ ✤â
❉♦
1
.
10x2 − 11x + 3
1
a + b + c = 7x − 4 +
10x2 − 11x + 3
1 1 1
1
1
x+y+z = + + =
+
+ 10x2 − 11x + 3.
a b c
2x − 1 5x − 3
✤â a + b + c = a1 + 1b + 1c t÷ì♥❣ ✤÷ì♥❣ ✈î✐
1
1
1
+
+ 10x2 − 11x + 3 = 7x − 4 +
2x − 1 5x − 3
10x2 − 11x + 3
1
1
1
⇔
+
+ 10x2 − 18x + 7 =
.
2x − 1 5x − 3
10x2 − 11x + 3
❱➟② t❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
●✐↔✐✳
1
1
1
=
+
+ 10x2 − 18x + 7.
2
10x − 11x + 3 2x − 1 5x − 3
✣✐➲✉ ❦✐➺♥ x = 53 , x = 12 . ✣➦t
1
a = 2x − 1, b = 5x − 3, c =
.
10x2 − 11x + 3
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✼✽
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❑❤✐ ✤â abc = 1 ✈➔ tø ✭✶✮ s✉② r❛ a + b + c = a1 + 1b + 1c ✳ ❚❛ ❝❤ù♥❣
♠✐♥❤ ✤÷ñ❝ ❦➳t q✉↔✿ ◆➳✉ xyz = 1 ✈➔ x + y + z = x1 + y1 + z1 t❤➻
(x − 1) (y − 1) (z − 1) = 0.
❚ø ✤â
❱➟②
1
(1) ⇔ (2x − 1 − 1) (5x − 3 − 1)
−1
10x2 − 11x + 3
x=1
x=1
4
4
x
=
⇔
⇔x=
5
√
5
11 ± 41
2
10x − 11x + 2 = 0
.
x=
20
√
4 11 ± 41
.
♣❤÷ì♥❣ tr➻♥❤ ❝â t➟♣ ♥❣❤✐➺♠ S = 1; 5 ; 20
=0
✹✳ ❉ò♥❣ ♠➺♥❤ ✤➲✿ ❱î✐ a, b, c ❧➔ ❝→❝ sè t❤ü❝ t❤ä❛ ♠➣♥ a + b + c = 0
t❤➻
1
1
1
1 1 1
+
+
=
+ + .
a2 b2 c2
a b c
❈❤ù♥❣ ♠✐♥❤✳ ❱î✐ a + b + c = 0 t❛ ❝â
2
1
1
1
1
1
1
1 1 1
+ +
= 2 + 2 + 2 +2
+ +
a b c
a
b
c
ab bc ac
1
1
1
2 (a + b + c)
1
1
1
= 2+ 2+ 2+
= 2 + 2 + 2.
a
b
c
abc
a
b
c
❱➟②
1
1
1
1 1 1
+
+
=
+ +
a2 b2 c2
a b c
✳
❇➔✐ t♦→♥✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
1
1
3
2 +
2 =
(2x − 1)
(3x + 1)
(x + 2)2
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x ∈/
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
1 −1
;
; −2
2 3
(1)
✳ ❚❛ ❝❤ù♥❣ ♠✐♥❤ ✤÷ñ❝ ❱î✐ a, b, c ❧➔
✼✾
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❝→❝ sè t❤ü❝ t❤ä❛ ♠➣♥ a + b + c = 0 t❤➻
❚❛ ❝â
1 1 1
1
1
1
+
+
=
+ + .
a2 b 2 c 2
a b c
(1) ⇔
1
1
1
3
.
2 +
2 +
2 =
(2x − 1)
(−3x − 1)
(x + 2)
(x + 2)2
(2)
❱➻ (2x − 1) + (3x + 1) + (x + 2) = 0 ♥➯♥
1
1
1
1
1
1
.
+
+
=
+
+
(2x − 1) (−3x − 1) (x + 2)
(2x − 1)2 (−3x − 1)2 (x + 2)2
❱➟② ✭✷✮ t÷ì♥❣ ✤÷ì♥❣ ✈î✐
1
1
1
2
+
+
=
(2x − 1) (−3x − 1) (x + 2)
|x + 2|
(−3x − 1) (x + 2) + (2x − 1) (x + 2) + (2x − 1) (−3x − 1)
2
=
(2x − 1) (−3x − 1) (x + 2)
|x + 2|
√
−7x2 − 3x − 3
1± 5
⇔
= 2 ⇔ x2 − x − 1 = 0 ⇔ x =
.
(2x − 1) (−3x − 1)
2
√
1± 5
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ♥❣❤✐➺♠ x = 2 .
⇔
✷✳✾ ▼ët sè ❤÷î♥❣ s→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾
✷✳✾✳✶ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ tø ❝→❝ ✤➥♥❣ t❤ù❝
❳✉➜t ♣❤→t tø ♠ët ✤➥♥❣ t❤ù❝ ♥➔♦ ✤â✱ ❝❤ó♥❣ t❛ ❝â t❤➸ s→♥❣ t→❝ ❧➯♥ ❝→❝
♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾✳ ❈❤➥♥❣ ❤↕♥ tø ❤➡♥❣ ✤➥♥❣ t❤ù❝
(a + b + c)3 = a3 + b3 + c3 + 3 (a + b) (b + c) (c + a)
t❛ ❝â
(a + b + c)3 = a3 + b3 + c3 ⇔ (a + b) (b + c) (c + a) = 0.
❇➡♥❣ ❝→❝❤ ❝❤å♥ a, b, c s❛♦ ❝❤♦ (a + b + c)3 = a3 + b3 + c3 t❛ s➩ t↕♦ r❛
✤÷ñ❝ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ ❝❤ù❛ ❝➠♥ ❜➟❝ ❜❛✳ ❙❛✉ ✤➙② t❛ s➩ s→♥❣ t→❝ ♠ët
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✽✵
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
sè ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ tø ❝→❝ ❤➡♥❣ ✤➥♥❣ t❤ù❝
(a + b + c)3 = a3 + b3 + c3 + 3 (a + b) (b + c) (c + a)
a3 + b3 − ab (a + b) = (a + b) (a − b)2 .
a3 + b3 = (a + b) a2 − ab + b2 .
a4 + 4 = a2 − 2a + 2 a2 + 2a + 2 .
√
√
√
❱î✐ a = 3 7x + 1, b = − 3 x2 − x − 8, c = 3 x2 − 8x − 1✱
❱➼ ❞ö ✶✳
t❛
❝â a3 + b3 + c3 = 8✳ ❚❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ✭✣➲ ♥❣❤à ❖❧②♠♣✐❝ ✸✵✴✵✹✴✶✾✾✾✮✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
3
●✐↔✐✳ ❚➟♣ ①→❝ ✤à♥❤
c=
√
3
x2 − 8x − 1✳
3
3
x2 − x − 8 + x2 − 8x − 1 = 2.
√
√
R✳ ✣➦t a = 3 7x + 1, b = − 3 x2 − x − 8✱
7x + 1 −
❑❤✐ ✤â
a3 + b 3 + c 3 = 8
(1)
a+b+c=2
(2)
▼➦t ❦❤→❝ t❛ ❝â ❤➡♥❣ ✤➥♥❣ t❤ù❝
(a + b + c)3 = a3 + b3 + c3 + 3 (a + b) (b + c) (c + a)
a = −b
❚❤❛② ✭✶✮✱ ✭✷✮ ✈➔♦ ✭✸✮ t❛ ✤÷ñ❝ (a + b) (b + c) (c + a) = 0 ⇔ b = −c .
c = −a
❱➟②
√
3
√
3
−x−8
7x + 1 = x2 − x − 8
√
√
3 x2 − x − 8 = 3 x2 − 8x − 1 ⇔ x2 − x − 8 = x2 − 8x − 1
√
√
3
3
x2 − 8x − 1 = − 7x + 1
x2 − 8x − 1 = −7x − 1
x = −1
x2 − 8x − 9 = 0
x=9
⇔ 7x = 7
⇔
x=1
x2 − x = 0
x = 0.
7x + 1 =
x2
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✽✶
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❚❤❛② ❝→❝ ❣✐→ trà −1, 0, 1, 9 ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t❤➜② t❤ä❛ ♠➣♥✳
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ✹ ♥❣❤✐➺♠ −1, 0, 1, 9✳
❱➼ ❞ö ✷✳ ❈❤♦ a = √1945x + 1975, b = √60x + 15, c = √15 − x t❤➻
3
3
3
t❛ ❝â a3 + b3 + c3 = 2004x + 2005✳ ❚❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
3
1945x + 1975 +
√
3
√
3
60x + 15 +
15 − x −
√
3
2004x + 2005 = 0
●✐↔✐✳ ❚➟♣ ①→❝ ✤à♥❤ R✳ ✣➦t
a=
√
3
1945x + 1975, b =
√
3
60x + 15, c =
√
3
15 − x.
❑❤✐ ✤â a3 + b3 + c3 = 2004x + 2005✳ ❚❤❛② ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t❛
✤÷ñ❝
a+b+c−
3
a3 + b3 + c3 = 0 ⇔ (a + b + c)3 = a3 + b3 + c3 .
(1)
▼➦t ❦❤→❝ t❛ ❧↕✐ ❝â ❤➡♥❣ ✤➥♥❣ t❤ù❝
(a + b + c)3 = a3 + b3 + c3 + 3 (a + b) (b + c) (c + a)
(2)
a = −b
❚ø ✭✶✮ ✈➔ ✭✷✮ s✉② r❛ (a + b) (b + c) (c + a) = 0 ⇔ b = −c . ❱➟②
c = −a
1990
x
=
−
1945x + 1975 = −60x − 15
2005
60x + 15 = x − 15
x = − 30
⇔
59
1990
15 − x = −(1945x + 1975)
x=−
1944
30
1990
❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ❜❛ ♥❣❤✐➺♠ x = − 1990
✱
x=− ✱x=−
✳
2005
59
1944
√
√
❈❤♦ a = 3 3x2 − x + 2001, b = − 3 3x2 − 7x + 2002,
√
c = − 3 6x − 2003 t❤➻ a3 + b3 + c3 = 2002✳ ❚❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❱➼ ❞ö ✸✳
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✽✷
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
3
3x2 − x + 2001 −
3
3x2 − 7x + 2002 −
√
3
6x − 2003 =
√
3
2002.
❍÷î♥❣ ❞➝♥✳ ✣➦t
a=
3
√
3
3x2 − x + 2001, b = − 3x2 − 7x + 2002, c = − 3 6x − 2003.
❑❤✐ ✤â (a + b + c)3 = a3 + b3 + c3 ⇔ (a + b) (b + c) (c + a) = 0✳ ❱✐➺❝
❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ✤÷ñ❝ q✉② ✈➲ ❣✐↔✐
√
3
3x2 − x + 2001 =
√
3
3x2 − 7x + 2002
√
3
3 3x2 − 7x + 2002 = −√
6x − 2003
√
√
3
6x − 2003 = 3 3x2 − x + 2001
❱➼ ❞ö ✹✳ ❚❛ s➩ s→♥❣ t→❝ ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❞ü❛ tr➯♥ ✤➥♥❣ t❤ù❝
a4 + 4 = a2 − 2a + 2 a2 + 2a + 2 .
√
√
√
√
❳➨t ✤➥♥❣ t❤ù❝ 2x4 + 4 = 2x2 + 2 4 2x + 2 2x2 − 2 4 3x + 2
√
u=
√
4
2x2 + 2 2x + 2, v =
√
✳ ✣➦t
√
4
2x2 − 2 2x + 2
√
❑❤✐ ✤â uv = 2x4 + 4✳ ◆➳✉ ♠✉è♥ s→♥❣ t→❝ ♠ët ♣❤÷ì♥❣ tr➻♥❤ ✤÷ñ❝
❣✐↔✐ ❜➡♥❣ ❝→❝❤ ✤÷❛ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ au2 − buv + cv2 = 0 t❛ ❝❤➾ ✈✐➺❝ t➼♥❤
au2 + cv 2 t❤❡♦ x✱ s❛✉ ✤â ❝❤♦ buv = au2 + cv 2 ✱ t❛ s➩ ✤÷ñ❝ ♠ët ♣❤÷ì♥❣
tr➻♥❤ ➞♥ x✳ ❳➨t
(3u − 2v) (2u − v) = 0 ⇔ 6u2 + 2v 2 = 7uv.
▼➔
6
√
√
√
√
√
√
2x2 + 2 4 2x + 2 + 2 2x2 − 2 4 2x + 2 = 8 2x2 + 8 4 2x + 16✱
♥➯♥ t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✹✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
√
√
8 2x2 + 8 4 2x + 16 = 7 2x4 + 4
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✽✸
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
●✐↔✐✳ √❚➟♣ ①→❝ √✤à♥❤ R✳ ✣➦t u =
√
√
2x2 + 2 4 2x + 2
2x2 − 2 4 2x + 2✳ ✣✐➲✉ ❦✐➺♥ u > 0, v > 0✳ ❑❤✐ ✤â
√
√
√
√
4
4
u2 = 2x2 + 2 2x + 2, v 2 = 2x2 − 2 2x + 2, u2 v 2 = 2x4 + 4,
√
√
6u2 + 2v 2 = 8 2x2 + 8 4 2x + 16 ❚❤❛② ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t❛ ✤÷ñ❝
2
u
=
v
3
6u2 + 2v 2 = 7uv ⇔ (3u − 2v) (2u − v) = 0 ⇔
v = 2u.
v=
•
❑❤✐ u = 32 v✱ t❛ ✤÷ñ❝
√
√
√
2 √ 2
4
4
2x2 + 2 2x + 2 =
2x − 2 2x + 2
3
√
√
√
4 √ 2
⇔ 2x2 + 2 4 2x + 2 =
2x − 2 4 2x + 2
9
√
√ 2
4
⇔ 5 2x + 26 2x + 10 = 0
√
−13 ± 119
√
⇔x=
542
•
❑❤✐ v = 2u✱ t❛ ✤÷ñ❝
√
❱➟②
√
√
√
2x2 − 2 4 2x + 2 = 2
2x2 + 2 4 2x + 2
√
√
√
√
⇔ 2x2 − 2 4 2x + 2 = 4 2x2 + 2 4 2x + 2
√
√
⇔ 3 2x2 + 10 4 2x + 6 = 0
√
−5 + 7
√
⇔x=
342
√
√
−13 ± 119
−5 + 7
√
√
♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ♥❣❤✐➺♠ x =
,x =
✳
542
342
❱➼ ❞ö ✺✳ ❚❛ s➩ s→♥❣ t→❝ ♠ët ♣❤÷ì♥❣ tr➻♥❤ ❞ü❛ tr➯♥ ❤➡♥❣ ✤➥♥❣ t❤ù❝
a3 + b3 = (a + b) a2 − ab + b2
√
❳➨t ✤➥♥❣ t❤ù❝ x3 + 83 = (x + 2) x2 − 2x + 4 . ✣➦t u = x + 2, v =
√
√
x2 − 2x + 4✳ ❑❤✐ ✤â uv = x3 + 83 ✳ ◆➳✉ ♠✉è♥ s→♥❣ t→❝ ♠ët ♣❤÷ì♥❣
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✽✹
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
tr➻♥❤ ✤÷ñ❝ ❣✐↔✐ ❜➡♥❣ ❝→❝❤ ✤÷❛ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ au2 − buv + cv2 = 0✱
t❛ ❝❤➾ ✈✐➺❝ t➼♥❤ au2 + cv2 t❤❡♦ x✱ t❛ s➩ ✤÷ñ❝ ♠ët ♣❤÷ì♥❣ tr➻♥❤ ➞♥ x✳
√
❈❤➥♥❣ ❤↕♥ t❛ ❝â 2u2 + v2 = x2 + 8✳ ❳➨t 2 2u2 + v2 = 5uv, t❛ ✤÷ñ❝
♣❤÷ì♥❣ tr➻♥❤
√
2 x2 + 8 = 5 x3 + 8
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✺✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x ≥ −2✳ ✣➦t
❦✐➺♥ u ≥ 0 ✈➔ v ≥ 0✳ ❑❤✐ ✤â
√
2 x2 + 8 = 5 x3 + 8
√
√
u = x + 2, v = x2 − 2x + 4✳
√
✣✐➲✉
u2 = x + 2, v 2 = x2 − 2x + 4, u2 v 2 = x3 + 8, 2u2 + v 2 = x2 + 8.
❚❤❛② ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ t❛ ✤÷ñ❝
√
•
√
2 2u2 + v 2 = 5uv ⇔ 2 2u − v
√
❑❤✐ 2
2u = v ✱
√ √
2 2 x+2=
√
• ❑❤✐ u = 2v ✱
u−
√
2v = 0 ⇔
√
2 2u = v
√
u = 2v.
t❛ ✤÷ñ❝
x2 − 2x + 4 ⇔ x2 − 10x − 12 = 0 ⇔ x = 5 ±
√ √
√
t❛ ✤÷ñ❝ x + 2 = 2 x2 − 2x + 4
√
37.
✭❱æ ♥❣❤✐➺♠✮
√
❑➳t ❤ñ♣ ✈î✐ ✤✐➲✉ ❦✐➺♥ t❛ ✤÷ñ❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❧➔ x = 5± 37✳
⇔ 2x2 − 5x + 6 = 0.
✷✳✾✳✷ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ tø ❝→❝ ❤➺ ✤è✐ ①ù♥❣ ❧♦↕✐ ■■
❳➨t ❤➺ ♣❤÷ì♥❣ tr➻♥❤
2
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
(αx + β) = ay + b
(1)
(αy + β)2 = ax + b
(2)
✽✺
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❚ø ✭✷✮ t❛ ❝â
√
ax + b β
−
ax + b
y = √α
α
⇔
√
− ax + b β
αx + β = − ay + b
y=
− .
α
α
αy + β =
√
❚❤➳ ✈➔♦ ✭✶✮ t❛ ✤÷ñ❝
√
a
ax + b aβ
2
(αx
+
β)
=
−
+ b (∗)
α
√α
a ax + b aβ
2
(αx + β) =
−
+ b.
α
α
✣➳♥ ✤➙②✱ ❜➡♥❣ ❝→❝❤ ❝❤å♥ α, β, a, b t❛ s➩ s→♥❣ t→❝ ✤÷ñ❝ ❝→❝ ♣❤÷ì♥❣ tr➻♥❤
√
✈æ t➾✳ ❈→❝❤ ❣✐↔✐ ❝→❝ ♣❤÷ì♥❣ tr➻♥❤ ❞↕♥❣ ♥➔② ❧➔ ✤➦t αy + β = ax + b
√
❤♦➦❝ αy = − ax + b ✤➸ ✤÷❛ ✈➲ ❤➺ ✤è✐ ①ù♥❣ ❧♦↕✐ ❤❛✐ ð tr➯♥✳ ❇➙② ❣✐í
t❛ s➩ ✤✐ s→♥❣ t→❝ ♠ët sè ♣❤÷ì♥❣ tr➻♥❤ ❞↕♥❣ ♥➔②✳
❱➼ ❞ö ✶✳ ❚❤❛② α = 2, β = 1, a = 2, b = 5 t❤❛② ✈➔♦ ✭✯✮ t❛ ✤÷ñ❝
(2x + 1)2 =
√
2x + 5 + 4.
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
4x2 + 4x − 3 = 3 2x + 5
. P❤÷ì♥❣ tr➻♥❤ ✈✐➳t ❧↕✐
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x ≥ −5
2
(2x + 1)2 − 4 =
✣➦t 2y + 1 =
r❛ t❛ ❝â ❤➺
√
2x + 5✱
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
√
2x + 5
(1)
s✉② r❛ (2y + 1)2 = 2x + 5 ❑➳t ❤ñ♣ ✈î✐ ✭✶✮ s✉②
(2x + 1)2 = 2y + 5
(2)
(2y + 1)2 = 2x + 5
(3)
✽✻
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
✣➯ x, y t❤ä❛ ♠➣♥ ✭✶✮ ✈➔ ✭✷✮ t❤➻ x ≥ − 52 ✈➔ y ≥ − 52 . ▲➜② ✭✷✮ trø ✭✸✮
t❛ ✤÷ñ❝
2 (x − y) (2x + 2y + 2) = 2 (y − x) ⇔ (x − y) (2x + 2y + 3) = 0
⇔
•
x−y =0
2x + 2y + 3 = 0
⇔
y=x
2y = −(2x + 3).
❱î✐ y = x✱ t❤❛② ✈➔♦ ✭✷✮ t❛ ✤÷ñ❝
−1 ±
(2x + 1)2 = 2x + 5 ⇔ 4x2 + 2x − 4 = 0 ⇔ x =
2
✤✐➲✉ ❦✐➺♥✮
• ❱î✐ y = x✱ t❤❛② ✈➔♦ ✭✷✮ t❛ ✤÷ñ❝
√
5
✭t❤ä❛ ♠➣♥
√
13
−3
±
✭t❤ä❛ ♠➣♥
(2x + 1) = −2x + 2 ⇔ 4x2 + 6x − 1 = 0 ⇔ x =
4
√
−1 ± 5
✤✐➲✉ ❦✐➺♥✮ ❈→❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❧➔ x = 2 ✈➔
√
−3 ± 13
x=
.
4
2
❱➼ ❞ö ✷✳ ❈❤♦ α = 1, β = 1, a = 21 , b = 32 t❤❛② ✈➔♦ ✭✯✮ t❛ ✤÷ñ❝
(x + 1)2 =
x 3
+
2 2 1 3
− + ⇔ 2(x + 1)2 =
2
2 2
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
2x2 + 4x =
x 3
+ + 2.
2 2
x+3
.
2
❱➼ ❞ö ✸✳ ❈❤♦ α = 2, β = −1, a = 8000, b = 1 t❤❛② ✈➔♦ ✭✯✮ t❛ ✤÷ñ❝
√
(2x − 1)2 = 4000 8000x + 1 + 4001.
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
x2 − x − 1000 8000x + 1 = 1000.
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✽✼
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
◆➳✉ ①➨t ❤➺
(αx + β)3 = ay + b
(αy + β)3 = ax + b
t❤➻ tø ♣❤÷ì♥❣ tr➻♥❤ ❞÷î✐ ✱ t❛ ✤÷ñ❝
αy + β =
3
√
3
αx + β ⇔ y =
❚❤❛② ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ tr➯♥ ❝õ❛ ❤➺ ✿
αx + β β
− .
α
α
√
a 3 αx + β aβ
(αx + β) =
−
+ b.
α
α
3
❱➼ ❞ö ✹✳ ❈❤å♥ α = 1, β = 1, a = 3, b = 5✱ t❛ ✤÷ñ❝
√
(x + 1)3 = 3 3 3x + 5 + 2.
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✹✳ ✭✣➲ ♥❣❤à ❖❧②♠♣✐❝ ✸✵✴✵✹✴✷✵✵✾✮✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
√
x3 + 3x2 − 3 3 3x + 5 = 1 − 3x.
❱➼ ❞ö ✺✳ ❈❤å♥ α = 2, β = 0, a = 4004, b = −2001✱ t❛ ✤÷ñ❝
√
(2x)3 = 2002 3 4004x − 2001 − 2001.
❚❛ ❝â ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✺✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
8x3 + 2001
2002
3
= 4004x − 2001.
✷✳✾✳✸ ❙→♥❣ t↕♦ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾ ❞ü❛ ✈➔♦ t➼♥❤ ✤ì♥ ✤✐➺✉ ❤➔♠ sè
❉ü❛ ✈➔♦ ❦➳t q✉↔✿ ✧◆➳✉ ❤➔♠ sè y = f (x) ✤ì♥ ✤✐➺✉ tr➯♥ ❦❤♦↔♥❣ (a; b)
t❤➻ f (x) = f (y) ⇔ x = y( ✈î✐x, y ∈ (a; b)) ✧✱ t❛ ❝â t❤➸ s→♥❣ t→❝ ✤÷ñ❝
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✽✽
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
♥❤✐➲✉ ♣❤÷ì♥❣ tr➻♥❤ ✈æ t➾✳
❱➼ ❞ö ✶✳ ❳➨t ❤➔♠ sè f (t) = t3 + 2t ✤ç♥❣ ❜✐➳♥ tr➯♥ R✳ ❈❤♦
f
3
−x3 + 9x2 − 19x + 11 = f (x − 1).
❚❛ ✤÷ñ❝
−x3 + 9x2 − 19x + 11 + 2 −x3 + 9x2 − 19x + 11 = (x − 1)3 + 2(x − 1).
3
❑❤❛✐ tr✐➸♥ ✈➔ rót ❣å♥ t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✶✳ ✭✣➲ ♥❣❤à ❖❧②♠♣✐❝ ✸✵✴✵✹✴✷✵✵✾✮✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
x3 − 6x2 + 12x − 7 = −x3 + 9x2 − 19x + 11.
√
✣➦t y = 3 −x3 + 9x2 − 19x + 11. ❚❛ ❝â ❤➺
3
●✐↔✐✳
y 3 = −x3 + 9x2 − 19x + 11
y = x3 − 6x2 + 12x − 7
⇔
y 3 = −x3 + 9x2 − 19x + 11
2y = 2x3 − 12x2 + 24x − 14
❈ë♥❣ ❤❛✐ ♣❤÷ì♥❣ tr➻♥❤ ✈î✐ ♥❤❛✉ t❛ ✤÷ñ❝
y 3 + 2y = x3 − 3x2 + 5x − 3 ⇔ y 3 + 2y = (x − 1)3 + 2 (x − 1)
✭✯✮.
❳➨t ❤➔♠ sè f (t) = t3 + 2t✳ ❱î✐ ♠å✐ t1 = t2✱ t❛ ❝â
f (t1 ) − f (t2 )
= t21 + t1 t2 + t22 + 2 =
t1 − t2
t2
t1 +
2
2
3t22
+
+ 2 > 0.
2
❱➟② ❤➔♠ sè f (t) ✤ç♥❣ ❜✐➳♥ tr➯♥ R✳ ❉♦ ✤â
(∗) ⇔ f (y) = f (x − 1) ⇔ y = x − 1
√
⇔ 3 −x3 + 9x2 − 19x + 11 = x − 1
⇔ −x3 + 9x2 − 19x + 11 = x3 − 3x2 + 3x − 1
x=1
⇔ x3 − 6x2 + 11x − 6 = 0 ⇔
x=2
x = 3.
P❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ❜❛ ♥❣❤✐➺♠ x = 1, x = 2, x = 3.
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✽✾
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❱➼ ❞ö ✷✳ ❳➨t ❤➔♠ sè f (t)
f (2x + 1) = f (−3x)✱
= t 2+
√
t2 + 3 .
t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❳➨t ♣❤÷ì♥❣ tr➻♥❤
❇➔✐ t♦→♥ ✷✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
3x 2 +
9x2 + 3 + (4x + 2)
●✐↔✐✳ ❚➟♣ ①→❝ ✤à♥❤ R✳ ❳➨t ❤➔♠
f (t) = 2 +
t2
x2 + x + 1 + 1 = 0.
√
sè f (t) = t 2 + t2 + 3
(1)
✳ ❚❛ ❝â
t2
+3+ √
> 0, ∀t ∈ R.
t2 + 3
❙✉② r❛ ❤➔♠ sè f ✤ç♥❣ ❜✐➳♥ tr➯♥ R✳ ❉♦ ✤â
(1) ⇔ (2x + 1) 2 +
(2x + 1)2 + 3
= (−3x) 2 +
(−3x)2 + 3
1
⇔ f (2x + 1) = f (−3x) ⇔ 2x + 1 = −3x ⇔ x = − .
5
❱➟② ✭✶✮ ❝â t➟♣ ♥❣❤✐➺♠ S = − 15 .
▲÷✉ þ✳ ◆❤➻♥ q✉❛ sü s➢♣ ①➳♣ ❝õ❛ ❜➔✐ t♦→♥✱√t❛ t❤➜② ✈➳ tr→✐ ❧➔ tê♥❣ ❝õ❛
❤❛✐ ❜✐➸✉ t❤ù❝ ❞↕♥❣ f (t) = (mt + n) p + at2 + bt + c ✳ ▼ët ❝→❝❤ tü
♥❤✐➯♥✱ t❛ ❤✐ ✈å♥❣ ❜✐➸✉ t❤ù❝ tr➯♥ ❝â t❤➸ ❝❤♦ t❛ ❞↕♥❣ ❝❤➼♥❤ t➢❝ ✤➸ ❞ò♥❣
✤ì♥ ✤✐➺✉✳ ✣➛✉ t✐➯♥ t❛ ✤÷❛ ♠é✐ ❜✐➸✉ t❤ù❝ ✈➲ ♠ët ✈➳✿
(1) ⇔ (4x + 2)
x2 + x + 1 + 1 = −3x 2 +
9x2 + 3 .
(2)
❚❛ ❦❤æ♥❣ t❤➸ ❝â 3x = m 9x2 + 3 + nx + p ✈➻ ❜✐➸✉ t❤ù❝ tr♦♥❣ ❝➠♥
❝â ❜➟❝ ❧î♥ ❤ì♥ ❜✐➸✉ t❤ù❝ ð ♥❣♦➔✐✳ ❱➟② t❛ s➩ ❧➔♠ ♥❣÷ñ❝ ❧↕✐✱ ♥❣❤➽❛
❧➔ ♣❤➙♥ t➼❝❤ ❜✐➸✉ t❤ù❝ ❜➟❝ ❧î♥ t❤❡♦ ❜✐➸✉ t❤ù❝ ❜➟❝ ♥❤ä✳ ❚❛ ❝â ✈➳
♣❤↔✐ ❝õ❛ ✭✷✮ ❜➡♥❣ −3x 2 + (−3x)2 + 3 ✳ ❚❛ ❤✐ ✈å♥❣✱ ✈➳ tr→✐ ❝õ❛
√
✭✷✮ ❝ô♥❣ ❝â t❤➸ ✤÷❛ ✈➲ ❞↕♥❣ f (t) = t 2 + t2 + 3 ✳ ▼ët ❝→❝❤ tü
♥❤✐➯♥✱ ✤➸ ①✉➜t ❤✐➺♥ sè ✷ tr♦♥❣ f (t) t❛ ❜✐➳♥ ✤ê✐ ✈➳ tr→✐ ❝õ❛ ✭✷✮ t❤➔♥❤
√
✿ (2x + 1) 2 + 4x2 + 4x + 4 ✳ ❉➵ t❤➜② 4x2 + 4x + 4 = (2x + 1)2 + 3✳
√
❱➟② t❛ ①➙② ❞ü♥❣ t❤➔♥❤ ❝æ♥❣ ❤➔♠ sè ✤ç♥❣ ❜✐➳♥ f (t) = t 2 + t2 + 3 ✳
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✾✵
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❱➼ ❞ö ✸✳ ❳➨t ❤➔♠ sè√ f (t) = t3 + t ✤ç♥❣ ❜✐➳♥ tr➯♥ R✱ rç✐ ①➨t ♣❤÷ì♥❣
tr➻♥❤ f (x + 1) = f
3x + 1
✱ t❛ ✤÷ñ❝ ❜➔✐ t♦→♥ s❛✉✳
❇➔✐ t♦→♥ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤
x3 + 3x2 + 4x + 2 = (3x + 2)
√
3x + 1
(1)
●✐↔✐✳ ✣✐➲✉ ❦✐➺♥ x ≥ − 13
√
(x + 1)3 + x + 1 = (3x + 1 + 1) 3x + 1
3
√
√
3x + 1 + 3x + 1
⇔ (x + 1)3 + x + 1 =
√
⇔ f (x + 1) = f
3x + 1 ✈î✐f (t) = t3 + t
√
⇔ x + 1 = 3x + 1 ✭❞♦ f (t) = t3 + t ✤ç♥❣ ❜✐➳♥✮
⇔ x2 + 2x + 1 = 3x + 1 ⇔
x=0
x=1
✭t❤ä❛ ✤✐➲✉ ❦✐➺♥✮
❱➟② ✭✶✮ ❝â t➟♣ ♥❣❤✐➺♠ S = {0; 1} .
▲÷✉ þ✿ ❚❤♦↕t ♥❤➻♥ ✈➳ tr→✐ ❝â ❜➟❝ ✸✱ ✈➳ ♣❤↔✐√❝â ❜➟❝ 32 ♥➯♥ ❦❤â ❝â t❤➸
❞ò♥❣ ✤ì♥ ✤✐➺✉✳ ◆❤÷♥❣ ♥➳✉ ✈➳ ♣❤↔✐ t❛ ❝♦✐ y = 3x + 1 ❧➔ ➞♥ t❤➻ ✈➳ ♣❤↔✐
❝ô♥❣ ❧➔ ❜➟❝ ❜❛ t❤❡♦ y✳ ❈ö t❤➸ ❝➛♥ ♣❤➙♥ t➼❝❤ 3x + 2 = m(3x + 1) + n✱
❦❤✐ ✤â ✈➳ ♣❤↔✐ ❝â ❞↕♥❣ my3 + ny. ❚❛ ❝â ♥❣❛② m = n = 1✳ ❈æ♥❣ ✈✐➺❝
❝á♥ ❧↕✐ ❧➔ ✤÷❛ ✈➳ tr→✐ ✈➲ ❞↕♥❣ (x − u)3 + x − u ❧➔ t❛ ❝â t❤➸ ❞ò♥❣ ✤ì♥
✤✐➺✉✳ ✣ç♥❣ ♥❤➜t ❤➺ sè t❛ ✤÷ñ❝ u = −1✳ ❱î✐ ♥❤ú♥❣ ❜➔✐ ♣❤÷ì♥❣ tr➻♥❤
t➼❝❤ ❝➛♥ ❧✐♥❤ ❤♦↕t tr♦♥❣ ✈✐➺❝ ✤ê✐ ❜✐➳♥ ✈➔ ①➙② ❞ü♥❣ ❤➔♠ ❝â t❤➸ ❝â ✤➸
❝â t❤➸ ✤÷❛ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ ❞↕♥❣ ❝❤➼♥❤ t➢❝✳ ▼ët sè ❜➔✐ ♥❤➻♥ ✈➔♦ r➜t
✧❦❤õ♥❣✧✱ ✤á✐ ❤ä✐ t❛ ♣❤↔✐ ❜➻♥❤ t➽♥❤ ♣❤➙♥ t➼❝❤✳ ❍➣② ♥❤î t❛ ❧✉æ♥ ❝è ❣➢♥❣
♣❤➙♥ t➼❝❤ ❜✐➸✉ t❤ù❝ ❜➟❝ ❧î♥ t❤❡♦ ❜✐➸✉ t❤ù❝ ❜➟❝ ♥❤ä✳
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✾✶
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❑➌❚ ▲❯❾◆
◆❤÷ ✈➟②✱ ✤➸ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ t❛ ❝â r➜t ♥❤✐➲✉ ♣❤÷ì♥❣ ♣❤→♣ ❦❤→❝
♥❤❛✉✳ ▼é✐ ♣❤÷ì♥❣ ♣❤→♣ ✤➲✉ ❝â ♥➨t ✤➦❝ tr÷♥❣ r✐➯♥❣ ❝õ❛ ♥â✳ ❚✉② ♥❤✐➯♥✱
✤➸ t➻♠ r❛ ♥❤ú♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ❤❛② ✈➔ ♥❣➢♥ ❣å♥ ✤á✐ ❤ä✐ ♥❣÷í✐ ❧➔♠
t♦→♥ ♣❤↔✐ ❝â ♥❤ú♥❣ ❤✐➸✉ ❜✐➳t ✤❛ ❞↕♥❣ ✈➲ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ✈î✐ ♠é✐
❧♦↕✐ ♣❤÷ì♥❣ tr➻♥❤✱ ♣❤↔✐ ❝â ♥❤ú♥❣ ❦✐➳♥ t❤ù❝ ❝ì ❜↔♥ t❤➟t ❝❤➢❝ ✈➔ ❦➽
♥➠♥❣ t➼♥❤ t♦→♥ t❤➔♥❤ t❤↕♦✳ ✣➸ r❛ ✤÷ñ❝ ♥❤ú♥❣ ✤➲ t♦→♥ ♠î✐ ✈➔ ❤❛②✱
♥❣÷í✐ r❛ ✤➲ ❝ô♥❣ ❝➛♥ ❝â ♥❤ú♥❣ ♣❤÷ì♥❣ ♣❤→♣ ①✉➜t ♣❤→t ♣❤ò ❤ñ♣ ✤➸
t↕♦ r❛ ♥❤ú♥❣ ♣❤÷ì♥❣ tr➻♥❤ ♥❤÷ ♠♦♥❣ ♠✉è♥✱ ♥❤➡♠ ❝õ♥❣ ❝è ❦✐➳♥ t❤ù❝
❝ì ❜↔♥ ✈➲ ♣❤÷ì♥❣ tr➻♥❤ ✈➔ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ t÷ì♥❣ ù♥❣ ❝õ❛ ❝❤ó♥❣✳
❚r♦♥❣ ❦❤â❛ ❧✉➟♥ ♥➔②✱ ❡♠ ✤➣ ✤÷❛ r❛ ♥❤ú♥❣ ❦✐➳♥ t❤ù❝ ❝ì ❜↔♥ ✈➲ ♣❤÷ì♥❣
tr➻♥❤✱ ♥❤ú♥❣ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ✈➔ ♠ët sè ❤÷î♥❣ ✤➸ t↕♦ r❛ ❝→❝ ✤➲ t♦→♥
♠î✐✳ ▼➦❝ ❞ò ♥❣♦➔✐ ♥❤ú♥❣ ♣❤÷ì♥❣ ♣❤→♣ ✤÷❛ r❛ ð ✤➙②✱ ❝á♥ r➜t ♥❤✐➲✉
♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ✈➔ s→♥❣ t↕♦ ✈î✐ ♥❤✐➲✉ ❞↕♥❣ ♣❤÷ì♥❣ tr➻♥❤ ✤❛ ❞↕♥❣
❦❤→❝✱ ♥❤÷♥❣ ❞♦ ❦❤✉æ♥ ❦❤ê ❝õ❛ ❦❤â❛ ❧✉➟♥ ✈➔ ❞♦ ♥➠♥❣ ❧ü❝ ❜↔♥ t❤➙♥ ❝á♥
♥❤✐➲✉ ❤↕♥ ❝❤➳ ♥➯♥ ❦❤â❛ ❧✉➟♥ ❝õ❛ ❡♠ ✈➝♥ ❝❤÷❛ ♥➯✉ ✤÷ñ❝ ✤➛② ✤õ ✈➔ ❤➺
t❤è♥❣ ❝→❝ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ✈➔ s→♥❣ t→❝ ♣❤÷ì♥❣ tr➻♥❤✳ ❍ì♥ ♥ú❛✱ ✤➙②
❧➔ ❧➛♥ ✤➛✉ t✐➯♥ ❧➔♠ q✉❡♥ ✈î✐ ♥❣❤✐➯♥ ❝ù✉ ❦❤♦❛ ❤å❝ ♥➯♥ tr♦♥❣ q✉→ tr➻♥❤
t❤ü❝ ❤✐➺♥ ✤➲ t➔✐✱ ❡♠ ❦❤æ♥❣ t❤➸ tr→♥❤ ❦❤ä✐ ♥❤ú♥❣ s❛✐ sât✳
❊♠ ❦➼♥❤ ♠♦♥❣ t❤➛② ❝æ ❣✐→♦✱ ❝→❝ ❜↕♥ s✐♥❤ ✈✐➯♥ ✤â♥❣ ❣â♣ t❤➯♠
þ ❦✐➳♥ ✤➸ ❜➔✐ ❦❤â❛ ❧✉➟♥ ❝õ❛ ❡♠ ✤÷ñ❝ ❤♦➔♥ t❤✐➺♥ ❤ì♥✳ ❊♠ ①✐♥ ❝❤➙♥
t❤➔♥❤ ❝↔♠ ì♥✦
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✾✷
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
❑❍➶❆ ▲❯❾◆ ❚➮❚ ◆●❍■➏P
❚❘×❮◆● ✣❍❙P ❍⑨ ◆❐■ ✷
❚⑨■ ▲■➏❯ ❚❍❆▼ ❑❍❷❖
❬✶❪ ◆❣✉②➵♥ ❚➔✐ ❈❤✉♥❣
❙→♥❣ t↕♦ ✈➔ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤✱ ❤➺ ♣❤÷ì♥❣ tr➻♥❤✱ ❜➜t ♣❤÷ì♥❣
tr➻♥❤✳ ◆❤➔ ①✉➜t ❜↔♥ tê♥❣ ❤ñ♣ t❤➔♥❤ ♣❤è ❍ç ❈❤➼ ▼✐♥❤ ✭✷✵✶✹✮✳
❬✷❪ ❚r➛♥ ❇→ ❍➔
❚rå♥❣ t➙♠ ❦✐➳♥ t❤ù❝ ✈➔ ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ t♦→♥ tr✉♥❣ ❤å❝ ♣❤ê
t❤æ♥❣ ✲ ✣↕✐ sè✳ ◆❤➔ ①✉➜t ❜↔♥ ❣✐→♦ ❞ö❝ ❱✐➺t ◆❛♠ ✭✷✵✵✾✮✳
❬✸❪ P❤❛♥ ❍✉② ❑❤↔✐
P❤÷ì♥❣ tr➻♥❤ ✈➔ ❜➜t ♣❤÷ì♥❣ tr➻♥❤ ✤↕✐ sè✳ ◆❤➔ ①✉➜t ❜↔♥ ❦❤♦❛
❤å❝ tü ♥❤✐➯♥ ✈➔ ❝æ♥❣ ♥❣❤➺ ✭✷✵✵✾✮✳
❬✹❪ ◆❣✉②➵♥ ❱➠♥ ▼➟✉
P❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ ✈➔ ❜➜t ♣❤÷ì♥❣ tr➻♥❤✳ ◆❤➔ ①✉➜t
❜↔♥ ❣✐→♦ ❞ö❝ ✭✷✵✶✵✮✳
❬✺❪ ❚r➛♥ P❤÷ì♥❣ ✲ ▲➯ ❍ç♥❣ ✣ù❝
❚✉②➸♥ t➟♣ ❝→❝ ❝❤✉②➯♥ ✤➲ ❧✉②➺♥ t❤✐ ✤↕✐ ❤å❝ ♠æ♥ t♦→♥ ✲ ✣↕✐ sè
sì ❝➜♣✳ ◆❤➔ ①✉➜t ❜↔♥ ❍➔ ◆ë✐ ✭✷✵✵✷✮✳
❬✻❪ ❚✉②➸♥ t➟♣ ✶✵ ♥➠♠ ✤➲ t❤✐ ❖❧②♠♣✐❝ ✸✵✴✵✹✳ ◆❤➔ ①✉➜t ❜↔♥ ❣✐→♦
❞ö❝ ✭✷✵✵✻✮✳
❑✸✼❇ ❙÷ ♣❤↕♠ ❚♦→♥
✾✸
❍♦➔♥❣ ❚❤ò② ▲✐♥❤
[...]... ì P Pữỡ số ỷ ử t t ừ số ữỡ tr ữỡ q tở õ ữợ ử s ữợ ỹ t ữợ ữợ ữỡ tr f (x) = k ữợ t số y = f (x) ũ số ỡ sỷ ỗ ữợ t ợ x = x0 f (x) = f (x0) = k õ x = x0 ợ x > x0 f (x) > f (x0) = k õ ữỡ tr ổ ợ x < x0 f (x) < f (x0) = k õ ữỡ tr ổ x0 t ừ ữỡ tr ữợ ỹ t ữợ ữợ ữỡ tr f (x) = g (x) ữợ t số y = f (x) y = g (x) ũ số y = f (x) ỗ ỏ số y = g (x) ... tr (0; +) tứ 2x 1 = 3(x 1)2 3x2 8x + 4 x ừ ữỡ tr S = ữ 2 3 2 2, 3 tọ 2, ũ P ì P Pữỡ số t t ữỡ tr x t số m f (x; m) = 0 tr ữỡ t m t số x m t x rỗ q x tữớ ũ ữỡ t số m õ t ợ t tự ừ ữỡ tr m õ số ữỡ ữỡ ừ ởt tự r ởt số trữớ ủ t ỏ õ t số ởt ữỡ rt t sỹ ừ ữỡ õ t ỳ ớ ở t t ữỡ tr (x + 1)2 + x + 6 = 5 Pữỡ tr t (1) (x + 1) + 5 = 5... ởt số ờ tữỡ ữỡ tữớ ữỡ tr f (x) = g (x) õ t D Rn số h(x) tr D t ữỡ tr (1) f (x) + h (x) = g (x) + h (x) số h(x) tr D t (1) f (x) h (x) = g (x) h (x) õ (1) (f (x))2n+1 = (g (x))2n+1, n N (1) (f (x))2n = (g (x))2n, n N f (x) g(x) ũ ữỡ ũ tr D t (1) (f (x))2n = (g (x))2n , n N ữ ũ P Pì PP Pì ữỡ ởt số ữỡ ữỡ tr Pữỡ ờ tữỡ ữỡ ú t sỷ ử ởt số. .. t số y = f (x) ũ số ỡ sỷ ỗ ữợ õ f (u) = f (v) u = v, u, v Df t ữỡ tr 3x + 1 + x + 7x + 2 = 4 ữ ũ P ì P ủ ỵ ợ t t tổ tữớ ữ ữỡ t ử t s õ t ỵ ởt út ú t s t tr ởt ỗ x = 1 ởt ừ ữỡ tr t t õ ữủ x = 1 t t õ ớ ữ s ớ D = x R|x t số f (x) = 3x + 1 + x + õ f tử tr 7 57 2 7x + 2 7 1+ 3 2 7x + 2 + f (x) = >0 2 3x + 1 2 x + 7x + 2 số. .. ờ t t (t 1) 2t2 4t + 3 = 0 t {0; 1} ứ õ s r x {0; 1} Pữỡ sỷ ử số t Pữỡ số t õ ú t t t ữủ ớ ữỡ tr r t s t ữỡ tổ q ởt số t s õ õ tr ỵ tữ ủ ỵ t ữỡtr 5 x(2x + 1) + 2x + 1 3 x = 8x + 1 ị tữ Pữỡ tr t ố rố tỹ ụ tứ ợt tự t s õ ởt s tỹ t s 8x + 1 t tự tr 2x + 1 x tốt t số t t s 8x + 1 = (2x + 1) + x 2 + = 8 =1 =1 =6 õ ớ x 0 õ... 2 + x4 4 17 = x 17x2 + 2 = 0 x2 = 2 x ữỡ tr õ ố x1,2 = x3,4 = ổ 17 3 2 17 3 2 17 3 2 ữ ũ P ì P t ữỡ tr x+ 11 + x = 11 ữ ỵ õ sỹ ừ số ụ ụ tứ ợ ỡ số õ sỹ ừ số ữợ tự õ t õ t ữỡ số t ó ớ s ớ 0 < x < 11 ợ t õ (1) 11 x = 11 + x (11 x)2 = 11 + x t 11 = a ữỡ tr ữủ t t (a x)2 = a + x a2 2ax + x2 = a + a2 (2x + 1) a + (x2 x) =... sỷ ử ỵ r ởt ữỡ t ợ t tớ ỡ s ừ ữỡ s ỵ r số y = f (x) tử tr [a; b] õ tr (a; b) t tỗ t ởt số c s f (b) f (a) = f (c) (b a) t s tr ởt ữỡ tr ữủ ử ỵ r ỏ ỳ ữ ữủ ú t r ữủ tũ ừ ỵ r ữủ tr ữỡ tr Pữỡ tr ah(x) bh(x) = k (a b) h (x) ợ 0 < a = 1 0 < b = 1a > b k 0 k = 1 h(x) tr [b; a] Pữỡ t t số t f (t) = th(x) k.h (x) t õ tứ t õ ah(x) kah(x) =... 3x + 2) = k x3 + 2x 2x3 x + 2 k = 1 ỏ ố ợ ữỡ tr tờ qt af (x) ag(x) = h (x) ữủ tữỡ tỹ ữớ t t ũ ữỡ số t ữ tr ữ h (x) = k [g (x) f (x)] ớ Pữỡ tr t 3 32x 32x x+2 3 x+2 3x 3 +2x = 2x3 x + 2 + x3 + 2x + 2x3 x + 2 = 3x 3 +2x + x3 + 2x f 2x3 x + 2 = f x3 + 2x ợ f (t) = 3t + t số f ỗ tr f (t) = 3t ln 3 + 1 > 0, t R ữ ũ P ì P (3) 2x3 x + 2 = x3 + 2x x3 3x + 2 =... (x) = g (x) f (x) = g 2 (x) ổ f (x) 0 f (x) 0 f (x)+ g (x) = h (x) g (x) 0 f (x) + g (x) + 2 f (x) g (x) = h (x) h (x) 0 ợ f (x) , g (x) , h (x) õ ổ ố ợ ữỡ tr rt số r ọ ữớ t õ t ụ õ t ũ ởt ỡ số ừ ữỡ tr ú t ữ ỵ ờ s loga f (x) = loga g (x) 0 0 P ì P loga f (x) = b f (x) = ab t ữỡ tr x 2x + 3 = 0 Pữỡ tr t ữợ x0 2x + 3 = x 2 2x + 3... (c) = 0 ứ t ữủ x s õ tỷ ồ t ữỡ tr t x = 0 ởt ừ sỷ ởt t ừ õ 3cos 2cos = cos 3cos 3 cos = ữ ũ 3cos x 2cos x = cos x P ì P 2cos x 2 cos t số f (t) = tcos t cos ợ t > 1 số f tử tr (1; +) õ f (t) = cos tcos 1 cos ứ t õ f (2) = f (3) f tử tr [2; 3] õ tr (2; 3) õ tỗ t b (2; 3) s cos 1 f (3) f (2) = f (b) (2 3) f (b) = 0 cos cos = 0 .b ... Pữỡ số t t ữỡ tr x t số m f (x; m) = tr ữỡ t m t số x m t x rỗ q x tữớ ũ ữỡ t số m õ t ợ t tự ữỡ tr m õ số ữỡ ữỡ ởt tự r ởt số trữớ ủ t ỏ õ t số ởt ữỡ... t số y = f (x) y = g (x) ũ số y = f (x) ỗ ỏ số y = g (x) x0 s f (x0) = g (x0) ữợ ữỡ tr õ t x = x0 ữợ ỹ t ữợ ữợ ữỡ tr f (u) = f (v) ữợ t số y = f (x) ũ số ... ự t Pữỡ s s t tờ ủ Pữỡ trú õ õ ỗ P tự ỡ P ởt số ữỡ ởt số s t r t ợ ữỡ tr ữỡ ởt số ữỡ ữỡ tr ữỡ ởt số s t r t ợ P tr ỳ tự ỡ ữỡ tr ỗ ữỡ tr t t ữỡ tr
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