❚❘×❮◆● ✣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