Đa thức bậc ba và các bài toán liên quan

BỘ GIÁO DỤC VÀ ĐÀO TẠO ĐẠI HỌC ĐÀ NẴNG NGUYỄN THỊ ĐẰM ĐA THỨC BẬC BA VÀ CÁC BÀI TOÁN LIÊN QUAN LUẬN VĂN THẠC SĨ KHOA HỌC Đà Nẵng - Năm 2015 BỘ GIÁO DỤC VÀ ĐÀO TẠO ĐẠI HỌC ĐÀ NẴNG NGUYỄN THỊ ĐẰM ĐA THỨC BẬC BA VÀ CÁC BÀI TOÁN LIÊN QUAN Chuyên ngành: Phương pháp Toán Sơ cấp Mã số: 60.46.01.13 LUẬN VĂN THẠC SĨ KHOA HỌC Người hướng dẫn khoa học: GS TSKH NGUYỄN VĂN MẬU Đà Nẵng - Năm 2015 ▲❮■ ❈❆▼ ✣❖❆◆ ❚æ✐ ❝❛♠ ✤♦❛♥ ✤➙② ❧➔ ❝æ♥❣ tr➻♥❤ ♥❣❤✐➯♥ ❝ù✉ ❝õ❛ r✐➯♥❣ tæ✐✳ ❈→❝ sè ❧✐➺✉✱ ❦➳t q✉↔ ♥➯✉ tr♦♥❣ ❧✉➟♥ ✈➠♥ ❧➔ tr✉♥❣ t❤ü❝ ✈➔ ❝❤÷❛ tø♥❣ ✤÷đ❝ ❛✐ ❝æ♥❣ ❜è tr♦♥❣ ❜➜t ❦➻ ❝æ♥❣ tr➻♥❤ ♥➔♦ ❦❤→❝✳ ❚→❝ ❣✐↔ ❧✉➟♥ ✈➠♥ ◆❣✉②➵♥ ❚❤à ✣➡♠ ▼Ö❈ ▲Ö❈ ▼Ð ✣❺❯ ỵ t➔✐ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ▼ö❝ ✤➼❝❤ ✈➔ ♥❤✐➺♠ ✈ö ♥❣❤✐➯♥ ❝ù✉ ✳ ✳ ✳ ✳ ✣è✐ t÷đ♥❣ ✈➔ ♣❤↕♠ ✈✐ ♥❣❤✐➯♥ ❝ù✉ ✳ ✳ ✳ ✳ ✳ P❤÷ì♥❣ ♣❤→♣ ♥❣❤✐➯♥ ❝ù✉ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ Þ ♥❣❤➽❛ ❦❤♦❛ ❤å❝ ✈➔ t❤ü❝ t✐➵♥ ❝õ❛ ✤➲ t➔✐ ❈➜✉ tró❝ ❧✉➟♥ ✈➠♥ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✶ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✶ ✶ ✷ ✷ ✷ ✷ ❈❍×❒◆● ✶✳ ❑■➌◆ ❚❍Ù❈ ❈❒ ❇❷◆ ❱➋ ✣❆ ❚❍Ù❈ ❇❾❈ ❇❆ ✹ ✶✳✶✳ P❍×❒◆● ❚❘➐◆❍ ❇❾❈ ❇❆ ❱❰■ ❍➏ ❙➮ ❚❍Ü❈ ✳ ✳ ✳ ✳ ✳ ✹ ✶✳✷✳ ❙Û ❉Ư◆● P❍×❒◆● ❚❘➐◆❍ ❇❾❈ ❇❆ ✣➎ ●■❷■ P❍×❒◆● ❚❘➐◆❍ ❇❾❈ ❇➮◆ ❚✃◆● ◗❯⑩❚ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✶✵ ❈❍×❒◆● ✷✳ ❍➏ P❍×❒◆● ❚❘➐◆❍ ❇❆ ❽◆ ✷✳✶✳ ▼❐❚ ❙➮ ✣➬◆● ◆❍❻❚ ❚❍Ù❈ ❇❆ ❇■➌◆ ✳ ✳ ✷✳✷✳ ❍➏ P❍×❒◆● ❚❘➐◆❍ ✣➮■ ❳Ù◆● ❇❆ ❽◆ ✳ ✳ ✷✳✷✳✶✳ ữỡ tr s ỵ t ♣❤÷ì♥❣ tr➻♥❤ ❤♦→♥ ✈à ✈á♥❣ q✉❛♥❤ ✳ ✷✳✸✳ ▼❐❚ ❙➮ P❍×❒◆● ❚❘➐◆❍ ✣×❆ ❱➋ ❍➏ ❇❆ ✶✸ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ❽◆ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✶✸ ✶✾ ✶✾ ✷✹ ✷✾ ❈❍×❒◆● ✸✳ P❍×❒◆● ❚❘➐◆❍ ❇❾❈ ❇❆ ▲■➊◆ ◗❯❆◆ ❚❆▼ ●■⑩❈ ✸✸ ✸✳✶✳ P❍×❒◆● ❚❘➐◆❍ ❇❾❈ ❇❆ ❚❍❊❖ ❈⑩❈ ❨➌❯ ❚➮ ❚❘❖◆● ❚❆▼ ●■⑩❈ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✸✳✶✳✶✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ t❤❡♦ ❝→❝ ❝↕♥❤ tr♦♥❣ t❛♠ ❣✐→❝ ✸✳✶✳✷✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ t❤❡♦ ❝→❝ ✤÷í♥❣ ❝❛♦ ✳ ✳ ✳ ✳ ✳ ✸✳✶✳✸✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ t❤❡♦ ❝→❝ ❜→♥ ❦➼♥❤ ✤÷í♥❣ trá♥ ❜➔♥❣ t✐➳♣ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✸✳✶✳✹✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ t❤❡♦ ❝→❝ ❤➔♠ ❧÷đ♥❣ ❣✐→❝ tr♦♥❣ t❛♠ ❣✐→❝ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✳ ✸✳✷✳ ❙Û ❉Ư◆● P❍×❒◆● ❚❘➐◆❍ ❇❾❈ ❇❆ ❑❍❷❖ ❙⑩❚ ❈⑩❈ ❇⑨■ ❚❖⑩◆ ❈Ü❈ ❚❘➚ ❚✃◆● ◗❯⑩❚ ❚❘❖◆● ❚❆▼ ●■⑩❈ ✸✸ ✸✸ ✸✺ ✸✻ ✸✼ ✹✽ ❑➌❚ ▲❯❾◆ ✺✻ ❚⑨■ ▲■➏❯ ❚❍❆▼ ❑❍❷❖ ✺✼ ◗❯❨➌❚ ✣➚◆❍ ●■❆❖ ✣➋ ❚⑨■ ▲❯❾◆ ❱❿◆ ✭❜↔♥ s❛♦✮ ◆❍Ú◆● ❑➑ ❍■➏❯ ❉Ò◆● ❚❘❖◆● ▲❯❾◆ ❱❿◆ ❑➼ ❤✐➺✉ ❚➯♥ ❣å✐ r R S, p a, b, c , hb , hc , rb , rc ✣ë ❞➔✐ ❜→♥ ❦➼♥❤ ✤÷í♥❣ trá♥ ♥ë✐ t✐➳♣ ❝õ❛ t❛♠ ❣✐→❝✳ ✣ë ❞➔✐ ❜→♥ ❦➼♥❤ ✤÷í♥❣ trá♥ ♥❣♦↕✐ t✐➳♣ ❝õ❛ t❛♠ ❣✐→❝✳ ❙è ✤♦ ❞✐➺♥ t➼❝❤ ✈➔ ♥û❛ ❝❤✉ ✈✐ ❝õ❛ t❛♠ ❣✐→❝✳ ✣ë ❞➔✐ ❝→❝ ❝↕♥❤ ❝õ❛ t❛♠ ❣✐→❝✳ ✣ë ❞➔✐ ❝→❝ ✤÷í♥❣ ❝❛♦ ❝õ❛ t❛♠ ❣✐→❝✳ ✣ë ❞➔✐ ❜→♥ ❦➼♥❤ ❝→❝ ✤÷í♥❣ trá♥ t t é ỵ ❞♦ ❝❤å♥ ✤➲ t➔✐ ✣❛ t❤ù❝ ❝â ✈à tr➼ r➜t q✉❛♥ trå♥❣ tr♦♥❣ ❚♦→♥ ❤å❝ ❦❤ỉ♥❣ ♥❤ú♥❣ ♥❤÷ ❧➔ ♠ët ✤è✐ t÷đ♥❣ ♥❣❤✐➯♥ ❝ù✉ trå♥❣ t➙♠ ❝õ❛ ✣↕✐ sè ♠➔ ❝á♥ ❧➔ ♠ët ❝ỉ♥❣ ❝ư ✤➢❝ ❧ü❝ ❝õ❛ ●✐↔✐ t➼❝❤ tr♦♥❣ ❧➼ t❤✉②➳t ①➜♣ ①➾✱ ❧➼ t❤✉②➳t ❜✐➸✉ ❞✐➵♥✱ ❧➼ t❤✉②➳t ✤✐➲✉ ❦❤✐➸♥✱ tè✐ ÷✉✳✳✳ ◆❣♦➔✐ r❛✱ ❧➼ t❤✉②➳t ✤❛ t❤ù❝ ❝á♥ ✤÷đ❝ sû ❞ư♥❣ ♥❤✐➲✉ tr♦♥❣ ❚♦→♥ ❝❛♦ ❝➜♣✱ ❚♦→♥ ù♥❣ ❞ö♥❣✳ ❚r♦♥❣ ❝→❝ ❦➻ t❤✐ ❤å❝ s✐♥❤ ❣✐ä✐ q✉è❝ ❣✐❛✱ ❖❧②♠♣✐❝ q✉è❝ t➳ t❤➻ ❝→❝ ❜➔✐ t♦→♥ ✈➲ ✤❛ t❤ù❝ ❝ơ♥❣ ✤÷đ❝ ✤➲ ❝➟♣ ♥❤✐➲✉ ✈➔ ✤÷đ❝ ①❡♠ ♥❤÷ ♥❤ú♥❣ ❞↕♥❣ t♦→♥ ❦❤â ❝õ❛ ❜➟❝ ♣❤ê t❤ỉ♥❣✳ ❈→❝ ❜➔✐ t♦→♥ ❧✐➯♥ q✉❛♥ ✤➳♥ ✤❛ t❤ù❝ ❝ô♥❣ ♥➡♠ tr♦♥❣ ❝❤÷ì♥❣ tr➻♥❤ t❤✐ ❖❧②♠♣✐❝ s✐♥❤ ✈✐➯♥ ❣✐ú❛ ❝→❝ tr÷í♥❣ ✣↕✐ ❤å❝ ✈➔ ❈❛♦ ✤➥♥❣ ✈➲ ●✐↔✐ t➼❝❤ ✈➔ ✣↕✐ sè✳ ❚r♦♥❣ ❝❤÷ì♥❣ tr➻♥❤ ♣❤ê t❤ỉ♥❣✱ ❝❤ó♥❣ t❛ ✤➣ ❧➔♠ q✉❡♥ ✈ỵ✐ ❦❤→✐ ♥✐➺♠ ✤❛ t❤ù❝ tø ❜➟❝ tr✉♥❣ ❤å❝ ❝ì sð✱ tø ♥❤ú♥❣ ♣❤➨♣ ❝ë♥❣✱ trø✱ ♥❤➙♥ ✤❛ t❤ù❝ ✤➳♥ t tự r tứ số ũ sỡ ỗ ❍♦r♥❡r ✤➸ ❝❤✐❛ ✤❛ t❤ù❝✱ ❣✐↔✐ ❝→❝ ♣❤÷ì♥❣ tr➻♥❤ ✤↕✐ sè✳ P❤➛♥ ✤↕✐ sè ❤➛✉ ❤➳t ✤➲✉ ♥❣❤✐➯♥ ❝ù✉ ✈➲ ✤❛ t❤ù❝ ❜➟❝ ♥❤➜t✱ ❜➟❝ ❤❛✐ ✈➔ ♠ët sè ✤❛ t❤ù❝ ❞↕♥❣ ✤➦❝ ❜✐➺t ❜➟❝ ❝❛♦✳ ❘➜t ♥❤✐➲✉ ù♥❣ ❞ö♥❣ ✈➔ ❜➔✐ t➟♣ ✤➣ ✤÷đ❝ ❤å❝ tr♦♥❣ ❝❤÷ì♥❣ tr➻♥❤ ♣❤ê t❤ỉ♥❣✳ ❱ỵ✐ ♠♦♥❣ ♠✉è♥ t➻♠ ❤✐➸✉ s➙✉ ✈➔ ❦➽ ❤ì♥ ✈➲ ✤❛ t❤ù❝ ❜➟❝ ❜❛ ✈➔ ♥➯✉ ❝→❝❤ ❣✐↔✐ q✉②➳t t q ữợ sỹ ữợ ●❙✳ ❚❙❑❍✳ ◆❣✉②➵♥ ❱➠♥ ▼➟✉ tæ✐ ✤➣ ❝❤å♥ ✤➲ t➔✐✿ ✏✣❛ t❤ù❝ ❜➟❝ ❜❛ ✈➔ ❝→❝ ❜➔✐ t♦→♥ ❧✐➯♥ q✉❛♥✑✳ ✷✳ ▼ö❝ ✤➼❝❤ ♥❣❤✐➯♥ ❝ù✉ ▼ö❝ ✤➼❝❤ ♥❣❤✐➯♥ ❝ù✉ ❝õ❛ ✤➲ t➔✐ ❧➔ t➻♠ ❤✐➸✉ ❝→❝ ❦✐➳♥ t❤ù❝ ❝ì ❜↔♥ ✈➲ ✤❛ t❤ù❝ ❜➟❝ ❜❛ ✈➔ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛✱ ✤❛ t❤ù❝ ❜➟❝ ❜❛ ❧✐➯♥ q✉❛♥ ✤➳♥ ❝→❝ ②➳✉ tè tr♦♥❣ t❛♠ ❣✐→❝✱ ❝→❝ ❞↕♥❣ ✤❛ t❤ù❝ ✤è✐ ①ù♥❣ ❜❛ ❜✐➳♥✱ ❤➺ ♣❤÷ì♥❣ tr➻♥❤ t✉②➳♥ t➼♥❤ ❜❛ ➞♥✳✳✳ ✷ ✸✳ ✣è✐ t÷đ♥❣ ✈➔ ♣❤↕♠ ✈✐ ♥❣❤✐➯♥ ❝ù✉ ✲ ✣❛ t❤ù❝ ❜➟❝ ❜❛ ✈➔ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛✳ ✲ ✣❛ t❤ù❝ ❜➟❝ ❜❛ ❧✐➯♥ q✉❛♥ ✤➳♥ t❛♠ ❣✐→❝✱ ❦❤↔♦ s→t ❝→❝ ❞↕♥❣ t♦→♥ ❝ü❝ trà ❞↕♥❣ ❦❤æ♥❣ ✤è✐ ①ù♥❣✳ ✲ ❈→❝ ❞↕♥❣ ✤❛ t❤ù❝ ✤è✐ ①ù♥❣ ❜❛ ❜✐➳♥✱ ❤➺ ♣❤÷ì♥❣ tr➻♥❤ t✉②➳♥ t➼♥❤ ❜❛ ➞♥✳✳✳ ✹✳ P❤÷ì♥❣ ♣❤→♣ ♥❣❤✐➯♥ ❝ù✉ ✲ ❚❤✉ t❤➟♣ ❝→❝ t➔✐ ❧✐➺✉ ✈➲ ✤❛ t❤ù❝✱ ✤❛ t❤ù❝ ❜➟❝ ❜❛✱ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ♥❤÷ s→❝❤ ❣✐→♦ ❦❤♦❛✱ s→❝❤ ❣✐→♦ ✈✐➯♥✱ ❣✐→♦ tr➻♥❤✱ ❝→❝ t➔✐ ❧✐➺✉ ❝❤✉②➯♥ ✤➲ ✈➲ ✤❛ t❤ù❝ ❜➟❝ ❜❛✱ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛✱ ✳ ✳ ✳ ✲ ❑❤↔♦ s→t✱ ♣❤➙♥ t➼❝❤✱ tê♥❣ ❤ñ♣ t➔✐ ❧✐➺✉ ✤➸ ❤➺ t❤è♥❣ ✈➔ ♣❤➙♥ ❧♦↕✐ ❝→❝ ❞↕♥❣ t♦→♥ ✈➲ ✤❛ t❤ù❝ ❜➟❝ ❜❛ ✈➔ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛✱ ❤➺ ♣❤÷ì♥❣ tr➻♥❤ t✉②➳♥ t➼♥❤ ❜❛ ➞♥✱ ✤❛ t❤ù❝ ✤è✐ ①ù♥❣ ❜❛ ❜✐➳♥✳ ✲ ❚r❛♦ ✤ê✐✱ t❤↔♦ ❧✉➟♥✱ t❤❛♠ ❦❤↔♦ ỵ ữợ ỗ t❤ü❝ ❤✐➺♥ ✤➲ t➔✐✳ ✺✳ Þ ♥❣❤➽❛ ❦❤♦❛ ❤å❝ ✈➔ t❤ü❝ t✐➵♥ ❝õ❛ ✤➲ t➔✐ ✲ ❍➺ t❤è♥❣ ✈➔ ♣❤➙♥ ❧♦↕✐ ❝→❝ ❞↕♥❣ t♦→♥ ✈➲ ✤❛ t❤ù❝ ❜➟❝ ❜❛ ✈➔ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛✱ ❤➺ ♣❤÷ì♥❣ tr➻♥❤ t✉②➳♥ t➼♥❤ ❜❛ ➞♥✱ ✤❛ t❤ù❝ ✈➔ ♣❤➙♥ t❤ù❝ ✤è✐ ①ù♥❣ ❜❛ ❜✐➳♥✳ ✲ ▲➔♠ t➔✐ ❧✐➺✉ ❝❤♦ ❝→❝ ♥❣❤✐➯♥ ❝ù✉ ✈➲ s❛✉✳ ✻✳ ❈➜✉ tró❝ ❧✉➟♥ ✈➠♥ ◆❣♦➔✐ ♣❤➛♥ ▼ð ✤➛✉ ✈➔ ❑➳t ❧✉➟♥✱ ❧✉➟♥ ✈➠♥ ✤÷đ❝ ❝❤✐❛ ❧➔♠ ❜❛ ❝❤÷ì♥❣✿ ❈❤÷ì♥❣ ✶✳ ❑✐➳♥ t❤ù❝ ❝ì ❜↔♥ ✈➲ ✤❛ t❤ù❝ ❜➟❝ ❜❛✳ Pữỡ tr ợ số tỹ ❙û ❞ư♥❣ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ✤➸ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜è♥ tê♥❣ q✉→t✳ ❈❤÷ì♥❣ ✷✳ ❍➺ ♣❤÷ì♥❣ tr➻♥❤ ❜❛ ởt số ỗ t tự ❍➺ ♣❤÷ì♥❣ tr➻♥❤ ✤è✐ ①ù♥❣ ❜❛ ➞♥✳ ✸ ✷✳✸✳ ▼ët sè ♣❤÷ì♥❣ tr➻♥❤ ✤÷❛ ✈➲ ❤➺ ❜❛ ➞♥✳ ❈❤÷ì♥❣ ✸✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ❧✐➯♥ q✉❛♥ t❛♠ ❣✐→❝✳ ✸✳✶✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ t❤❡♦ ❝→❝ ②➳✉ tè tr♦♥❣ t❛♠ ❣✐→❝✳ ✸✳✷✳ ❙û ❞ư♥❣ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ❦❤↔♦ s→t ❝→❝ ❜➔✐ t♦→♥ ❝ü❝ trà tê♥❣ q✉→t tr♦♥❣ t❛♠ ❣✐→❝✳ ✹ ❈❍×❒◆● ✶ ❑■➌◆ ❚❍Ù❈ ❈❒ ❇❷◆ ❱➋ ✣❆ ❚❍Ù❈ ❇❾❈ ❇❆ ❚r♦♥❣ ❝❤÷ì♥❣ ♥➔②✱ ❝❤ó♥❣ tỉ✐ tr➻♥❤ ❜➔② ♠ët sè ❦✐➳♥ t❤ù❝ ❝ì ❜↔♥ ✈➲ ✤❛ t❤ù❝ ❜➟❝ ❜❛ ❞ü❛ tr➯♥ t➔✐ ❧✐➺✉ t❤❛♠ ❦❤↔♦ ❬✷❪ ✈➔ ❬✻❪✳ ✶✳✶✳ P❍×❒◆● ❚❘➐◆❍ ❇❾❈ ❇❆ ❱❰■ ❍➏ ❙➮ ❚❍Ü❈ ▼➦❝ ❞ò ❝→❝❤ ữỡ tr tờ qt ổ ữủ ợ t❤✐➺✉ ð ❜➟❝ ♣❤ê t❤ỉ♥❣✱ ♥❤÷♥❣ ❝→❝ ❜➔✐ t♦→♥ ❧✐➯♥ q✉❛♥ ✤➳♥ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ❧↕✐ t❤÷í♥❣ ❣➦♣ tr♦♥❣ ❝→❝ ❦➻ t❤✐ ✈➔♦ ✣↕✐ ❤å❝ ✈➔ t❤✐ ❤å❝ s✐♥❤ ọ r tr ỵ t t ỵ t ữỡ tr tờ qt ỵ t t ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ x1 , x2 , x3 ax3 + bx2 + cx + d = (a = 0) t❤➻  b   x + x + x = −    ca x1 x2 + x2 x3 + x3 x1 =  a   d  x1 x2 x3 =− a ỵ t số  u + v + w uv + vw + wu  uvw t❤➻ ❜❛ sè u, v, w ◆➳✉ x+y+z     xy + yz + zx    xyz t❤ä❛ ♠➣♥ =S =P = Q ❧➔ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❇➔✐ t♦→♥ ✶✳✶✳ ●✐↔✐❤➺ ♣❤÷ì♥❣ tr➻♥❤ u, v, w ❧➔ ❝→❝ ♥❣❤✐➺♠ x3 − Sx2 + P x − Q = =0 =− = ✹✸ A B C , cos2 , cos2 ❧➔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ 2 4R + r p2 + (4R + r)2 p2 t − t + t− = 2R 16R2 16R2 1 ❇➔✐ t♦→♥ ✸✳✸✽✳ ❈❤ù♥❣ ♠✐♥❤ r➡♥❣ , , ❧➔ ❝→❝ A B C 2 cos cos cos 2 ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❱➟② cos2 t3 − p2 + (4R + r)2 8R(4R + r) 16R2 t + t − = p2 p2 p2 ❈❤ù♥❣ ♠✐♥❤✳ ♣❤↔✐ ❝❤ù♥❣ ♠✐♥❤✳ ❚❤❛② t ❜ð✐ ✈➔♦ ♣❤÷ì♥❣ tr➻♥❤ ✭✸✳✹✻✮ t❛ ❝â ✤✐➲✉ t ❡✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ t❤❡♦ cot A, cot B, cot C ❇➔✐ t♦→♥ ✸✳✸✾✳ ❈❤ù♥❣ ♠✐♥❤ r➡♥❣ cot A, cot B, cot C ❧➔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ p2 − r2 − 4Rr (2R + r)2 − p2 t − t +t+ = 2pr 2pr ✭✸✳✹✼✮ ❚❤❡♦ ✤à♥❤ ❧➼ ❤➔♠ sè ❝♦s✐♥ t❛ ❝â✿ 2bc cos A = b2 + c2 − a2 ❤❛② 2bc sin A cot A = b2 + c2 − a2 ❤❛② ❈❤ù♥❣ ♠✐♥❤✳ b2 + c2 − a2 4S a2 + c − b a2 + b2 − c2 ❚÷ì♥❣ tü✿ cot B = , cot C = 4S 4S 2 a +b +c ❤❛② cot A + cot B + cot C = 4pr p2 − r2 − 4Rr ❙✉② r❛ cot A + cot B + cot C = 2pr cos A cos B cos C cot A cot B cot C = sin A sin B sin C ❱➟② ♥➯♥ 4S cot A = b2 + c2 − a2 ✈➔ cot A = p2 − (2R + r)2 p2 − (2R + r)2 4R cot A cot B cot C = = pr 2pr 2R2 ✹✹ ❱➻ A + B + C = π ⇒ B + C = π − A ♥➯♥ cot B cot C − cot(B + C) = cot(π − A) = − cot A ✈➔ = − cot A cot B + cot C ⇔ cot A cot B + cot B cot C + cot C cot A = ❉♦ ✤â cot A, cot B, cot C ❧➔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ p2 − r2 − 4Rr (2R + r)2 − p2 t − t +t+ = 2pr 2pr ❇➔✐ t♦→♥ ✸✳✹✵✳ ❈❤ù♥❣ ♠✐♥❤ r➡♥❣ tan A, tan B, tan C ❧➔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ 2pr p2 − r2 − 4Rr 2pr t − t + t − = p − (2R + r)2 p2 − (2R + r)2 p2 − (2R + r)2 1 ♥➯♥ ❦❤✐ t❤❛② t ❜ð✐ ✈➔♦ ♣❤÷ì♥❣ cot A t tr➻♥❤ ✭✸✳✹✼✮ t❛ ❝â ✤✐➲✉ ♣❤↔✐ ❝❤ù♥❣ ♠✐♥❤✳ A B C A ❢✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ t❤❡♦ tan , tan , tan ✈➔ cot , 2 2 C B cot , cot 2 A B C ❇➔✐ t♦→♥ ✸✳✹✶✳ ❈❤ù♥❣ ♠✐♥❤ r➡♥❣ tan , tan , tan ❧➔ ❝→❝ ♥❣❤✐➺♠ 2 ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ 4R + r r t3 − t + t − = ✭✸✳✹✽✮ p p ❈❤ù♥❣ ♠✐♥❤✳ ❉♦ tan A = ❈❤ù♥❣ ♠✐♥❤✳ ❚ø ❝æ♥❣ t❤ù❝ t➼♥❤ ❞✐➺♥ t➼❝❤ S = pr ✈➔ r = (p − a) tan A B C = (p − b) tan = (p − c) tan 2 s✉② r❛ A r B r C r tan = , tan = , tan = ❤❛② p−a p−b p−c A B C r r r r3 tan tan tan = = 2 p − a p − b p − c (p − a)(p − b)(p − c) pr3 pr3 = = p(p − a)(p − b)(p − c) S pr3 r = 2= pr p ✹✺ tan A B C r r r + tan + tan = + + 2 p−a p−b p−c 1 + + =r p−a p−b p−c ❱➟② ♥➯♥ tan A B C 4R + r 4R + r + tan + tan = r = 2 pr p A B B C C A tan + tan tan + tan tan 2 2 2 r r r r r r = + + p − ap − b p − bp − c p − cp − a 1 1 1 = r2 + + p − ap − b p − bp − c p − cp − a ❙✉② r❛ tan tan B B C C A A tan + tan tan + tan tan = r2 = 2 2 2 r ❚ø ✤â tan A B C , tan , tan ❧➔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ 2 t3 − 4R + r r t + t − = p p A ❇➔✐ t♦→♥ ✸✳✹✷✳ ❈❤ù♥❣ ♠✐♥❤ r➡♥❣ cot , cot ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ B C , cot ❧➔ ❝→❝ ♥❣❤✐➺♠ 2 p 4R + r p t3 − t2 + t − = r r r ❈❤ù♥❣ ♠✐♥❤✳ ❉♦ cot A = A tr➻♥❤ ✭✸✳✹✽✮ t❛ ❝â ✤✐➲✉ ♣❤↔✐ ❝❤ù♥❣ ♠✐♥❤✳ tan ♥➯♥ t❤❛② t ❜ð✐ ✈➔♦ ♣❤÷ì♥❣ t ❇➡♥❣ ❝→❝❤ t÷ì♥❣ tü t❛ ❝ơ♥❣ ①➙② ❞ü♥❣ ✤÷đ❝ ❝→❝ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ s❛✉✿ ✹✻ A B C , tan2 , tan2 2 A B C ❇➔✐ t♦→♥ ✸✳✹✸✳ ❈❤ù♥❣ ♠✐♥❤ r➡♥❣ tan2 , tan2 , tan2 ❧➔ ❝→❝ 2 ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❣✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ❝õ❛ tan2 r2 2p2 − (4R + r)2 p2 − 2r2 − 8Rr t + t − = t + p2 p2 p ✭✸✳✹✾✮ ❚❛ ❝â A B C (4R + r)2 − 2p2 tan2 + tan2 + tan2 = 2 p2 r2 A B C tan tan tan = 2 2 p p2 − 2r2 − 8Rr B B C C A A tan + tan tan + tan tan = tan 2 2 2 p2 A B C ❱➟② tan2 , tan2 , tan2 ❧➔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ 2 r2 2p2 − (4R + r)2 p2 − 2r2 − 8Rr t + t − = t + p2 p2 p ❈❤ù♥❣ ♠✐♥❤✳ A B C , cot2 , cot2 2 A B C ❇➔✐ t♦→♥ ✸✳✹✹✳ ❈❤ù♥❣ ♠✐♥❤ r➡♥❣ cot2 , cot2 , cot2 ❧➔ ❝→❝ 2 ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❤✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ❝õ❛ cot2 p2 p2 − 2r2 − 8Rr 2p2 − (4R + r)2 t + t − = r2 r2 r2 1 ❈❤ù♥❣ ♠✐♥❤✳ ❉♦ tan α = ✱ ♥➯♥ ❦❤✐ t❤❛② t ❜ð✐ ✈➔♦ ♣❤÷ì♥❣ cot α t tr➻♥❤ ✭✸✳✹✾✮ t❛ ✤÷đ❝ t3 − t 1+ 2p2 − (4R + r)2 + p2 t p2 − 2r2 − 8Rr r2 + − = ♥➯♥ p2 t p2 2p2 − (4R + r)2 p2 − 2r2 − 8Rr r2 t + t − t = ❤❛② p2 p2 p p2 − 2r2 − 8Rr 2p2 − (4R + r)2 p2 t − t + t − = r2 r2 r ✹✼ ✐✳ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ❝õ❛ a sin A, b sin B, c sin C ❇➔✐ t♦→♥ ✸✳✹✺✳ ❈❤ù♥❣ ♠✐♥❤ r➡♥❣ a sin A, b sin B, c sin C ❧➔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ p2 − r2 − 4Rr t + t − R ❈❤ù♥❣ ♠✐♥❤✳ p2 + r2 + 4Rr 2R 4rp2 2p2 r2 − t− = R R ❚❛ ❝â✿ a sin A + b sin B + c sin C a b c =a +b +c = (a + b2 + c2 ) 2R 2R 2R 2R p2 − r2 − 4Rr 2p2 − 2r2 − 8Rr = = 2R R a sin Ab sin B + b sin Bc sin C + c sin Ca sin A a b b c c a =a b +b c +c a 2R 2R 2R 2R 2R 2R 2 (a b + b2 c2 + c2 a2 ) = 4R ❙✉② r❛✿ a sin Ab sin B + b sin Bc sin C + c sin Ca sin A 2 = (p (p + r2 − 8Rr) + r2 (4R + r)2 ) 4R = ((p2 + r2 + 4R)2 − 8p2 Rr) 4R a sin Ab sin Bc sin C a b c a2 b c (4pRr)2 2p2 r2 =a b c = = = 2R 2R 2R 8R3 8R3 R ❚ø ✤â a sin A, b sin B, c sin C ❧➔ ❝→❝ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ p2 − r2 − 4Rr t − t + R p2 + r2 + 4Rr 2R 4rp2 2p2 r2 − t− = R R ✹✽ ✸✳✷✳ ❙Û ❉Ư◆● P❍×❒◆● ❚❘➐◆❍ ❇❾❈ ❇❆ ❑❍❷❖ ❙⑩❚ ❈⑩❈ ❇⑨■ ❚❖⑩◆ ❈Ü❈ ❚❘➚ ❚✃◆● ◗❯⑩❚ ❚❘❖◆● ❚❆▼ ●■⑩❈ ❇➔✐ t♦→♥ ✸✳✹✻✳ ●✐↔ sû ABC t❤❛② ✤ê✐✳ ✣➦t x = sin A, y = sin B, z = sin C ❍➣② t➻♠ ❣✐→ trà ❧ỵ♥ ♥❤➜t ❝õ❛ ❜✐➸✉ t❤ù❝ s❛✉ ✤➙②✿ T = 4(xy + yz + zx) + (x + y + z) ❇➔✐ ❣✐↔✐✳ ❱➻ a, b, c ❧➔ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ x3 − 2px2 + (p2 + r2 + 4Rr)x − 4Rrp = ♥➯♥ ab + bc + ca + (a + b + c)R = 2R2 + 4pR + p2 + r2 + 4Rr √ B C A 3R ♥➯♥ ❱➻ a + b + c = 8R cos cos cos 2 √ ab + bc + ca + (a + b + c)R (11 + 3)R2 √ ❤❛② 4(xy + yz + zx) + (x + y + z) 11 + √ ❱➟② ❣✐→ trà ❧ỵ♥ ♥❤➜t ❝õ❛ ❚ ❜➡♥❣ 11 + ❦❤✐ ABC ✤➲✉✳ ❇➔✐ t♦→♥ ✸✳✹✼✳ ❈❤♦ ABC ợ ữớ trỏ t R ❦➼♥❤ ❝→❝ ✤÷í♥❣ trá♥ ❜➔♥❣ t✐➳♣ ❧➔ r1 , r2 , r3 ✱ ♥û❛ ❝❤✉ ✈✐ p✳ ❈❤ù♥❣ ♠✐♥❤ r➡♥❣ r12 −1 p2 ❇➔✐ ❣✐↔✐✳ r22 −1 p2 r32 (2R + r)2 −1 =4 −1 p2 p2 ❱➻ r1 , r2 , r3 ❧➔ ❜❛ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ x3 − (4R + r)x2 + p2 x − p2 r = ❳→❝ ✤à♥❤ ♣❤÷ì♥❣ tr➻♥❤ ♥❤➟♥ y1 = r12 − p2 , y2 = r22 − p2 , y3 = r32 − p2 ❧➔♠ ❜❛ ♥❣❤✐➺♠✳ ❑❤û x tø ❤➺ ♣❤÷ì♥❣ tr➻♥❤ x3 − (4R + r)x2 + p2 x − p2 r = x2 − p2 − y = ✹✾ ❚❛ ❝â ♥❣❛② ❤➺ ♣❤÷ì♥❣ tr➻♥❤ x3 − (4R + r)x2 + p2 x − p2 r = x3 − p2 x − yx = ✈➔ (4R + r)x2 − yx − 2p2 x + p2 r = ❤❛② ♣❤÷ì♥❣ tr➻♥❤ (4R + r)(y + p2 ) − (y + 2p2 )x + p2 r = T y + (T + r)p2 ✣➦t T = 4R + r ❑❤✐ ✤â x = y + p2 (T y + (T + r)p2 )2 ❱➟② − p2 − y = ❤❛② 2 (y + 2p ) y(y + 2p2 )2 + p2 (y + 2p2 )2 − (T y + (T + r)p2 )2 = P❤÷ì♥❣ tr➻♥❤ ♥➔② ❝â ❜❛ ♥❣❤✐➺♠ ❧➔ y1 , y2 , y3 (r12 − p2 )(r22 − p2 )(r32 − p2 ) y1 y2 y3 (4R + 2r)2 p4 − 4p6 ❉♦ ✤â = = p6 p6 p6 ❚ø ❤➺ t❤ù❝ ố ũ t s r ữủ ỗ t tự r12 −1 p2 r22 −1 p2 r32 (2R + r)2 −1 =4 −1 p2 p2 ❇➔✐ t♦→♥ ✸✳✹✽✳ ❈❤♦ ABC ợ ữớ trỏ t r, R✱ ❜→♥ ❦➼♥❤ ❝→❝ ✤÷í♥❣ trá♥ ❜➔♥❣ t✐➳♣ ❧➔ r1 , r2 , r3 ❑❤✐ ✤â t❛ ❧✉æ♥ ❝â✿ 1 2R − r ✭✐✮ + + = (r − r1 )(r − r2 ) (r − r2 )(r − r3 ) (r − r3 )(r − r1 ) 2Rr2 1 1 ✭✐✐✮ + + − (r − r1 )(r − r2 ) (r − r2 )(r − r3 ) (r − r3 )(r − r1 ) r2 R2 ✭✐✐✐✮ ✣➦t T = 1 + + (r − r1 )(r − r2 ) (r − r2 )(r − r3 ) (r − r3 )(r − r1 ) ❳→❝ ✤à♥❤ ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ T ❦❤✐ r ❦❤æ♥❣ ✤ê✐✳ ❇➔✐ ❣✐↔✐✳ ❱➻ r1 , r2 , r3 ❧➔ ❜❛ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ x3 − (4R + r)x2 + p2 x − p2 r = ♥➯♥ ✺✵ f (x) = x3 − (4R + r)x2 + p2 x − p2 r = (x − r1 )(x − r2 )(x − r3 )✳ ▲➜② ❤❛✐ ❧➛♥ ✤↕♦ ❤➔♠ ✤÷đ❝ 3x − 4R − r = (x − r1 ) + (x − r2 ) + (x − r3 )✳ ❚ø ✤➙② ✤÷đ❝ 1 3x − 4R − r = + + f (x) (x − r1 )(x − r2 ) (x − r2 )(x − r3 ) (x − r3 )(x − r1 ) ❈❤♦ x = r ❝â 1 2R − r + + = (r − r1 )(r − r2 ) (r − r2 )(r − r3 ) (r − r3 )(r − r1 ) 2Rr2 1 2R − r − ♥➯♥ ❝â ✭✐✐✮ ✈➔ ✈➻ ❱➟② ❝â ✭✐✮ ✈➔ ❞♦ ❜ð✐ 2Rr2 r R2 2R − r 1 = − 2Rr2 r2 2Rr 4r2 ✳ ❉➜✉ ✧❂✧ ①↔② r❛ ❦❤✐ ABC ✤➲✉ ✈➔ ♥❤÷ t❤➳ ❣✐→ trà ♥❤ä ♥➯♥ T 4r2 ♥❤➜t ❝õ❛ T ❜➡♥❣ ✳ 4r ❇➔✐ t♦→♥ ✸✳✹✾✳ ❈❤♦ ABC ✈ỵ✐ ❜→♥ ❦➼♥❤ ✤÷í♥❣ trá♥ ♥ë✐ t✐➳♣ ❧➔ r ✈➔ ❝❤✉ ✈✐ 2p = 6✳ ❚➻♠ ❣✐→ trà ❧ỵ♥ ♥❤➜t ❝õ❛ ❜✐➸✉ tự ữợ T = 1 + + (4 − a)(4 − b) (4 − b)(4 − c) (4 − c)(4 − a) ❇➔✐ ❣✐↔✐✳ ❱➻ a, b, c ❧➔ ❜❛ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ x3 − 2px2 + (p2 + r2 + 4Rr)x − 4Rrp = ♥➯♥ f (x) = x3 − 6x2 + (9 + r2 + 4Rr)x − 12Rr = (x − a)(x − b)(x − c) ▲➜② ❤❛✐ ❧➛♥ ✤↕♦ ❤➔♠ ✤÷đ❝ 3x − 2p = (x − a) + (x − b) + (x − c) ❚ø ✤➙② ✤÷đ❝ 3x − 2p 1 = + + f (x) (x − a)(x − b) (x − b)(x − c) (x − c)(x − a) ✺✶ ❈❤♦ x = ❝â✿ 1 + + = (4 − a)(4 − b) (4 − b)(4 − c) (4 − c)(4 − a) + 2r2 + 2Rr ❱➻ R 2r s✉② r❛ 1 + + (4 − a)(4 − b) (4 − b)(4 − c) (4 − c)(4 − a) ✳ 6r2 + ❉➜✉ ✧❂✧ ①↔② r❛ ❦❤✐ R = 2r ❤❛② ABC ✤➲✉✳ ❉♦ ✈➟② ❣✐→ trà ❧ỵ♥ ♥❤➜t ❝õ❛ T ❧➔ ✳ ❇➔✐ t♦→♥ ✸✳✺✵✳ ❈❤♦ ABC ✈ỵ✐ ✤ë ❞➔✐ ❜❛ ❝↕♥❤ a, b, c ✈➔ ❜→♥ ❦➼♥❤ ✤÷í♥❣ trá♥ ♥ë✐ t✐➳♣ ❧➔ r✳ ❍➣② ①→❝ ✤à♥❤ ❣✐→ trà ❧ỵ♥ t tự ữợ p2 (a b)2 (b − c)2 (c − a)2 + + − T = ab bc ca 4r ❇➔✐ ❣✐↔✐✳ ❱➻ a, b, c ❧➔ ❜❛ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ x3 − 2px2 + (p2 + r2 + 4Rr)x − 4Rrp = ♥➯♥ ❞➵ ❞➔♥❣ ❝â (a − b)2 (b − c)2 (c − a)2 2(p2 + r2 + 4Rr) + + = − 9✳ ab bc ca 4Rr ◆❤÷ ✈➟② (a − b)2 (b − c)2 (c − a)2 2(p2 + r2 ) + + = − 7✳ ab bc ca 4Rr ❇ð✐ ✈➻ R 2r ♥➯♥ (a − b)2 (b − c)2 (c − a)2 + + ab bc ca 2(p2 + r2 ) p2 27 − = − 8r2 4r2 27 ❉♦ ✈➟② ❣✐→ trà ❧ỵ♥ ♥❤➜t ❝õ❛ T ❜➡♥❣ − ❦❤✐ ABC ✤➲✉✳ t ABC ợ ữớ trỏ ♥ë✐ t✐➳♣ ❧➔ r ❦❤æ♥❣ ✤ê✐ ✈➔ ✤ë ❞➔✐ ❝↕♥❤ a, b, c✳ ❍➣② ①→❝ ✤à♥❤ ❣✐→ trà ❧ỵ♥ ♥❤➜t ❝õ❛ ❜✐➸✉ t❤ù❝✿ 1 T = + + a b c ❇➔✐ ❣✐↔✐✳ ❱➻ a, b, c ❧➔ ❜❛ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✺✷ x3 − 2px2 + (p2 + r2 + 4Rr)x − 4Rrp = ♥➯♥ f (x) = x3 − 2px2 + (p2 + r2 + 4Rr)x − 4Rrp = (x − a)(x − b)(x − c) ▲➜② ❤❛✐ ❧➛♥ ✤↕♦ ❤➔♠ ✤÷đ❝ 3x − 2p = (x − a) + (x − b) + (x − c) ❚ø ✤➙② ✤÷đ❝ 1 3x − 2p = + + f (x) (x − a)(x − b) (x − b)(x − c) (x − c)(x − a) ❈❤♦ x = p ❝â 1 p + + = = (p − a)(p − b) (p − b)(p − c) (p − c)(p − a) r p r ❚â♠ ❧↕✐ 1 1 = + + 4r2 4(p − a)(p − b) 4(p − b)(p − c) 4(p − c)(p − a) ❉♦ ✈➟② ❣✐→ trà ❧ỵ♥ ♥❤➜t ❝õ❛ T ❜➡♥❣ ❦❤✐ 4r2 1 + + a2 b c ABC ✤➲✉✳ t sỷ ABC t ỵ ✤ë ❞➔✐ ❜❛ ❝↕♥❤ a, b, c ✈➔ ❜→♥ ❦➼♥❤ ✤÷í♥❣ trá♥ ♥ë✐✱ ♥❣♦↕✐ t✐➳♣ ❧➔ r, R✳ ❍➣② ①→❝ tr ợ t tự ữợ T = ❇➔✐ ❣✐↔✐✳ ab + bc + ca + (a + b + c)r + r2 R2 ❱➻ a, b, c ❧➔ ❜❛ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ x3 − 2px2 + (p2 + r2 + 4Rr)x − 4Rrp = ♥➯♥ 1 3x2 − 4px + p2 + r2 + 4Rr + + = x−a x−b x−c (x − a)(x − b)(x − c) ❱ỵ✐ x = −r ❝â 1 4r2 + 4pr + p2 + 4Rr + + = a+r b+r c+r (r + a)(r + b)(r + c) ❚ø ✤➙② s✉② r❛ ❜➜t ✤➥♥❣ t❤ù❝ ✺✸ √ 39 + 3 R2 (r + a)(r + b)(r + c) 1 + + a+r b+r c+r ❤❛② ab + bc + ca + (a + b + c)r + r2 ❱➟② ❣✐→ trà ❧ỵ♥ ♥❤➜t ❝õ❛ T ❜➡♥❣ √ 39 + 3 R2 √ 39 + 3 ❦❤✐ ABC ✤➲✉✳ ❇➔✐ t♦→♥ ✸✳✺✸✳ ❑➼ ❤✐➺✉ r1 , r2 , r3 ❧➔ ❜→♥ ❦➼♥❤ ❝→❝ ✤÷í♥❣ trá♥ ❜➔♥❣ t✐➳♣ ABC ✳ ❑❤✐ R ❦❤ỉ♥❣ t❤❛② ✤ê✐✱ ❤➣② ①→❝ ✤à♥❤ ❣✐→ trà ❧ỵ♥ ♥❤➜t ❝õ❛ tê♥❣ T = r1 r2 + r2 r3 + r3 r1 ❇➔✐ ❣✐↔✐✳ ❱➻ r1 , r2 , r3 ❧➔ ❜❛ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ x3 − (4R + r)x2 + p2 x − p2 r = ♥➯♥ r1 r2 + r2 r3 + r3 r1 = p2 ❱➻ p = R(sin A + sin B + sin C) √ 3R A+B+C = 3R sin ♥➯♥ r1 r2 + r2 r3 + r3 r1 27R2 27R2 ❱➟② ❣✐→ trà ❧ỵ♥ ♥❤➜t ❝õ❛ T ❜➡♥❣ ❦❤✐ ABC ✤➲✉✳ ❇➔✐ t♦→♥ ✸✳✺✹✳ ❈❤♦ ABC ợ ữớ trỏ t R ❜→♥ ❦➼♥❤ ❝→❝ ✤÷í♥❣ trá♥ ❜➔♥❣ t✐➳♣ ❧➔ r1 , r2 , r3 , ♥û❛ ❝❤✉ ✈✐ p✳ ❍➣② ①→❝ ✤à♥❤ ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ t➼❝❤ T = ❇➔✐ ❣✐↔✐✳ r12 +1 p2 r22 +1 p2 r32 +1 p2 r1 , r2 , r3 ❧➔ ❜❛ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✺✹ x3 − (4R + r)x2 + p2 x − p2 r = 0✳ ❳→❝ ✤à♥❤ ♣❤÷ì♥❣ tr➻♥❤ ♥❤➟♥ y1 = r12 + p2 , y2 = r22 + p2 , y3 = r32 + p2 ❧➔♠ ❜❛ ♥❣❤✐➺♠✳ ❑❤û x tø ❤➺ ♣❤÷ì♥❣ tr➻♥❤ x3 − (4R + r)x2 + p2 x − p2 r = x2 + p2 − y = ❚❛ ❝â ♥❣❛② ❤➺ ♣❤÷ì♥❣ tr➻♥❤ x3 − (4R + r)x2 + p2 x − p2 r = x3 + p2 x − yx = ✈➔ (4R + r)x2 − yx + p2 r = ❤❛② ♣❤÷ì♥❣ tr➻♥❤ (4R + r)(y − p2 ) − yx + p2 r = ✣➦t T = 4R + r ❑❤✐ ✤â x = T y − 4Rp2 ❱➟② y (T y − 4Rp2 )2 + p2 − y = y ❤❛② y − (T + p2 )y + 8RT p2 y − 16R2 p4 = P❤÷ì♥❣ tr➻♥❤ ♥➔② ❝â ❜❛ ♥❣❤✐➺♠ ❧➔ y1 , y2 , y3 (r12 + p2 )(r22 + p2 )(r32 + p2 ) y1 y2 y3 16R2 p4 ❉♦ ✤â = = p6 p6 p6 ❚ø ❤➺ t❤ù❝ ❝✉è✐ ũ t s r ữủ ỗ t tự r12 +1 p2 r22 +1 p2 ❱➟② ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ T ❧➔ r32 16R2 +1 = p2 p 64 ❦❤✐ 27 ABC ✤➲✉✳ 64 27 ✺✺ ❇➔✐ t♦→♥ ABC ợ ữớ trỏ t✐➳♣ ❧➔ r✱ ❜→♥ ❦➼♥❤ ❝→❝ ✤÷í♥❣ trá♥ ❜➔♥❣ t✐➳♣ ❧➔ r1 , r2 , r3 , ♥û❛ ❝❤✉ ✈✐ p✳ ❍➣② ①→❝ ✤à♥❤ ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ r12 + r22 + r32 ✭✐✮ T = p2 r12 + r22 + r32 ✭✐✐✮ T = r2 ❇➔✐ ❣✐↔✐✳ ✭✐✮ ❱➻ r1 , r2 , r3 ❧➔ ❜❛ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ x3 − (4R + r)x2 + p2 x − p2 r = ♥➯♥ r12 + r22 + r32 p2 r1 r2 + r2 r3 + r3 r1 = 1✳ p2 ❱➟② ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ T ❜➡♥❣ ✶ ❦❤✐ ✭✐✐✮ ❉♦ ❜ð✐ r12 + r22 + r32 r2 ABC ✤➲✉✳ 1 (r + r + r ) = (4R + r)2 2 3r 3r ♥➯♥ P 27✳ ❉♦ ✤â ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ P ❜➡♥❣ ✷✼ ❦❤✐ (8r + r)2 3r ABC ✤➲✉✳ ✺✻ ❑➌❚ ▲❯❾◆ ◆❤ú♥❣ ❦➳t q✉↔ ❝❤➼♥❤ ✤÷đ❝ tr➻♥❤ ❜➔② tr♦♥❣ ❧✉➟♥ ✈➠♥ ✏✣❛ t❤ù❝ ❜➟❝ ❜❛ ✈➔ ❝→❝ ❜➔✐ t♦→♥ ❧✐➯♥ q ỗ tố ❞↕♥❣ t♦→♥ ✈➲ ✤❛ t❤ù❝ ❜➟❝ ❜❛ ✈➔ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛✱ ❤➺ ♣❤÷ì♥❣ tr➻♥❤ t✉②➳♥ t➼♥❤ ❜❛ ➞♥✱ ✤❛ t❤ù❝ ✤è✐ ①ù♥❣ ❜❛ ❜✐➳♥✳ ✷✳ ❚r➻♥❤ ❜➔② ♠ët ❧ỵ♣ ❝→❝ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ♠➔ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❧➔ ❝→❝ ②➳✉ tè tr♦♥❣ t❛♠ ❣✐→❝✳ ✸✳ ❙û ❞ö♥❣ ♣❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ❦❤↔♦ s→t ❝→❝ ❜➔✐ t♦→♥ ❝ü❝ trà tê♥❣ q✉→t tr♦♥❣ t❛♠ ❣✐→❝✳ ❱ỵ✐ ♥❤ú♥❣ ❣➻ ✤➣ t➻♠ ❤✐➸✉ ✤÷đ❝✱ ❝❤ó♥❣ tỉ✐ ❤② ✈å♥❣ ❧✉➟♥ ✈➠♥ s➩ ❧➔ ♠ët t➔✐ ❧✐➺✉ t❤❛♠ ❦❤↔♦ ❤ú✉ ➼❝❤ ❝❤♦ ❜↔♥ t❤➙♥ tr♦♥❣ ❝æ♥❣ t→❝ ❣✐↔♥❣ ❞↕② ✈➔ ❤② ✈å♥❣ ❧✉➟♥ ụ ỗ tữ tốt s ♣❤ê t❤ỉ♥❣ ❝ơ♥❣ ♥❤÷ ♥❤ú♥❣ ❛✐ q✉❛♥ t➙♠ ✤➳♥ ✈➜♥ ✤➲ ♥➔②✳ ▼➦❝ ❞ị ✤➣ ❤➳t sù❝ ❝è ❣➢♥❣✱ ♥❤÷♥❣ ❞♦ t❤í✐ ❣✐❛♥ ✈➔ ❦❤↔ ♥➠♥❣ ❝â ❤↕♥ ♥➯♥ ❝❤➢❝ ❝❤➢♥ ❧✉➟♥ ✈➠♥ ❝á♥ ❝â ♥❤ú♥❣ t❤✐➳✉ sât✳ ❱➻ t❤➳ ú tổ rt ữủ ỵ õ õ qỵ t ổ ỗ ❧✉➟♥ ✈➠♥ ✤÷đ❝ ❤♦➔♥ t❤✐➺♥ ❤ì♥✳ ✺✼ ❚⑨■ ▲■➏❯ ❚❍❆▼ ❑❍❷❖ ❬✶❪ ❚r➛♥ ◆❛♠ ❉ơ♥❣ ✭✷✵✶✶✮✱ P❤÷ì♥❣ tr➻♥❤ ✈➔ ❤ì♥ t ỳ ố ỗ ❬✷❪ ◆❣✉②➵♥ ❱➠♥ ▼➟✉ ✭✷✵✵✷✮✱ ✣❛ t❤ù❝ ✤↕✐ sè ✈➔ ♣❤➙♥ t❤ù❝ ❤ú✉ t✛✱ ◆❳❇ ●✐→♦ ❞ö❝✳ ❬✸❪ ◆❣✉②➵♥ ❱➠♥ ▼➟✉ ✭✷✵✵✷✮✱ ▼ët sè ♣❤÷ì♥❣ ♣❤→♣ ❣✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ ✈➔ ❜➜t ♣❤÷ì♥❣ tr➻♥❤✱ ◆❳❇ ●✐→♦ ❞ư❝✳ ❬✹❪ ◆❣✉②➵♥ ❱➠♥ ▼➟✉✱ ▲➯ ◆❣å❝ ▲➠♥❣✱ P❤↕♠ ❚❤➳ ▲♦♥❣✱ ◆❣✉②➵♥ ▼✐♥❤ ❚✉➜♥ ✭✷✵✵✻✮✱ ❈→❝ ✤➲ t❤✐ ❖❧②♠♣✐❝ ❚♦→♥ s✐♥❤ ✈✐➯♥ t♦➔♥ q✉è❝✱ ◆❳❇ ●✐→♦ ❞ö❝✳ ❬✺❪ ◆❣✉②➵♥ ❱➠♥ ▼➟✉✱ P❤↕♠ ❇↕❝❤ ◆❣å❝ ✭✶✾✾✾✮✱ ▼ët sè ❜➔✐ t♦→♥ ❝❤å♥ ❧å❝ ✈➲ ❧÷đ♥❣ ❣✐→❝✱ ◆❳❇ ●✐→♦ ❞ö❝✳ ❬✻❪ ◆❣✉②➵♥ ❱➠♥ ▼➟✉✱ ◆❣✉②➵♥ ❱➠♥ ◆❣å❝ ✭✷✵✵✾✮✱ ✣❛ t❤ù❝ ✤è✐ ①ù♥❣ ✈➔ →♣ ❞ö♥❣✱ ◆❳❇ ●✐→♦ ❞ö❝✳ ỗ t t❤ù❝ ✈➔ ♣❤÷ì♥❣ ♣❤→♣ tå❛ ✤ë tr♦♥❣ ❤➻♥❤ ❤å❝✳ ❬✽❪ ◆❣✉②➵♥ ❱➠♥ ▼➟✉✱ ✣➦♥❣ ❍✉② ❘✉➟♥✱ ◆❣✉②➵♥ ▼✐♥❤ ❚✉➜♥ ✭✷✵✵✽✮✱ ❑✛ ②➳✉ tr↕✐ ❤➧ ❍ị♥❣ ❱÷ì♥❣ ❧➛♥ t❤ù ■❱✱ ❙ð ●❉ ✲ ✣❚ ❍á❛ ❇➻♥❤✳ ❬✾❪ ❚↕ ❉✉② P❤÷đ♥❣ ✭✷✵✵✹✮✱ P❤÷ì♥❣ tr➻♥❤ ❜➟❝ ❜❛ ✈➔ ❝→❝ ❤➺ t❤ù❝ tr♦♥❣ t❛♠ ❣✐→❝✱ ◆❳❇ ●✐→♦ ❞ö❝✳ ...BỘ GIÁO DỤC VÀ ĐÀO TẠO ĐẠI HỌC ĐÀ NẴNG NGUYỄN THỊ ĐẰM ĐA THỨC BẬC BA VÀ CÁC BÀI TỐN LIÊN QUAN Chun ngành: Phương pháp Tốn Sơ cấp Mã số: 60.46.01.13... a)(p − b)(p − c) 1+ =1+ = R pabc abc 2abc + (a + b − c)(b + c − a)(c + a − b) = 2abc 2 ab + ac + ba + bc2 + cb2 + ca2 − a3 − b3 − c3 = 2abc 2 2 a(b + c − a ) + b(a + c2 − b2 ) + c(a2 + b2 − c2