Xin giới thiệu tới các bạn học sinh, sinh viên Đề thi thử Đại học năm 2015 môn Toán - Đề số 16. Đề thi gồm có 10 câu hỏi tự luận có kèm đáp án và hướng dẫn giải chi tiết với thời gian làm bài 180 phút. Cùng tìm hiểu để nắm bắt nội dung thông tin tài liệu.
❚❍■ ❚❍Û ✣❸■ ❍➴❈ ◆❿▼ ✷✵✶✺ ✯✯✯✯✯✯✯✯✯✯ ✣➋ ❙➮ ✶✻ ▼ỉ♥✿ ❚♦→♥✳ ❚❤í✐ ❣✐❛♥✿ ✶✽✵ ♣❤ót ❈➙✉ ✶ ✭✷✱✵ ✤✐➸♠✮✳ ❈❤♦ ❤➔♠ sè x3 − 6x2 + 3(4 − m2)x + 6m2 − ✭✶✮✱ tr♦♥❣ ✤â m ❧➔ t❤❛♠ số st sỹ t ỗ t❤à ❝õ❛ ❤➔♠ sè ✭✶✮ ❦❤✐ m = 1✳ ❜✮ m ỗ t số t ỹ trà t↕✐ A✱ B s❛♦ ❝❤♦ t❛♠ ❣✐→❝ OAB ✈✉æ♥❣ t↕✐ O✳ ❈➙✉ ✷ ✭✶✱✵ ✤✐➸♠✮✳ ❛✮ ●✐↔✐ ❜➜t ♣❤÷ì♥❣ tr➻♥❤ 52x+1 − 26 · 5x + > (x ∈ R)✳ ❜✮ ❚➻♠ t➟♣ ❤ñ♣ ❝→❝ ✤✐➸♠ M ❜✐➸✉ ❞✐➵♥ sè ♣❤ù❝ z ❜✐➳t |z − 1| = |(1 − i)z|✳ ❈➙✉ ✸ ✭✶✱✵ ✤✐➸♠✮✳ ❚➼♥❤ t➼❝❤ ♣❤➙♥ √ √ 2 ex + x + + 2x 2x · I= ex ❞x ❈➙✉ ✹ ✭✶✱✵ ✤✐➸♠✮✳ ❚r♦♥❣ ❦❤ỉ♥❣ ❣✐❛♥ ✈ỵ✐ ❤➺ tå❛ ✤ë Oxyz✱ ❝❤♦ ♠➦t ♣❤➥♥❣ (P ) : x − 2y + 2z − = ✈➔ ❤❛✐ ✤÷í♥❣ −3 z x−5 y z+5 t❤➥♥❣ d1 : x −2 = y−3 = ✱ d2 : = = ✳ −5 ❛✮ ❱✐➳t ♣❤÷ì♥❣ tr➻♥❤ ♠➦t ♣❤➥♥❣ (Q) q✉❛ ✤÷í♥❣ t❤➥♥❣ d1 ✈➔ ✈✉ỉ♥❣ ❣â❝ ✈ỵ✐ ♠➦t ♣❤➥♥❣ (P )✳ ❜✮ ❚➻♠ ✤✐➸♠ M t❤✉ë❝ ✤÷í♥❣ t❤➥♥❣ d1 ✈➔ ✤✐➸♠ N t❤✉ë❝ ✤÷í♥❣ t❤➥♥❣ d2 s❛♦ ❝❤♦ ✤÷í♥❣ t❤➥♥❣ M N s s ợ t (P ) ỗ tớ ✤÷í♥❣ t❤➥♥❣ M N ❝→❝❤ ♠➦t ♣❤➥♥❣ (P ) ♠ët ❦❤♦↔♥❣ ❜➡♥❣ ✷✳ ❈➙✉ ✺ ✭✶✱✵ ✤✐➸♠✮✳ ❍➻♥❤ ❝❤â♣ S.ABC ❝â ✤→② ABC ❧➔ t❛♠ ❣✐→❝ ✈✉æ♥❣ t↕✐ B✱ BC = SA = a✱ ACB = SAC = 600✱ ♠➦t ♣❤➥♥❣ (SAC) ✈✉ỉ♥❣ ❣â❝ ✈ỵ✐ ♠➦t ✤→②✳ ❚➼♥❤ t❤❡♦ a t❤➸ t➼❝❤ ❦❤è✐ ❝❤â♣ S.ABC ✈➔ ❦❤♦↔♥❣ ❝→❝❤ tø ✤✐➸♠ A ✤➳♥ ♠➦t ♣❤➥♥❣ (SBC)✳ ❈➙✉ ✻ ✭✶✱✵ ✤✐➸♠✮✳ ❛✮ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ sin 3x + cos 2x = sin x, x ∈ (−π; π)✳ ❜✮ ▼ët ❧ỵ♣ ❝â ✸✵ ❤å❝ s✐♥❤ tr♦♥❣ ✤â ❝â ✸ ❤å❝ s✐♥❤ ❧➔ ❝→♥ ❜ë ❧ỵ♣✳ ❈❤å♥ ♥❣➝✉ ♥❤✐➯♥ ✸ ❤å❝ s✐♥❤ tø ❧ỵ♣ ✤➸ ❧➔♠ trü❝ ♥❤➟t ❧ỵ♣ ❤å❝✳ ❚➼♥❤ ①→❝ s✉➜t ✤➸ tr♦♥❣ ✸ ❤å❝ s✐♥❤ ✤÷đ❝ ❝❤å♥ ❝â ➼t ♥❤➜t ♠ët ❝→♥ ❜ë ❧ỵ♣✳ ❈➙✉ ✼ ✭✶✱✵ ✤✐➸♠✮✳ ❚r♦♥❣ ♠➦t ♣❤➥♥❣ ✈ỵ✐ ❤➺ tå❛ ✤ë Oxy✱ ❝❤♦ ❤➻♥❤ ✈✉ỉ♥❣ ABCD ❝â ✤✐➸♠ B t❤✉ë❝ ✤÷í♥❣ t❤➥♥❣ d : x + 3y − = 0✱ ✤✐➸♠ E t❤✉ë❝ t✐❛ ✤è✐ ❝õ❛ t✐❛ BA s❛♦ ❝❤♦ BA = 2BE ✱ ✤✐➸♠ H(7; 3) ❧➔ ❤➻♥❤ ❝❤✐➳✉ ✈✉æ♥❣ ❣â❝ ❝õ❛ A tr➯♥ CE ✱ ❤❛✐ ✤÷í♥❣ t❤➥♥❣ DH ✈➔ AB ❝➢t ♥❤❛✉ t↕✐ ✤✐➸♠ F (16; 0)✳ ❚➻♠ tå❛ ✤ë ❝→❝ ✤➾♥❤ ❝õ❛ ❤➻♥❤ ✈✉æ♥❣ ABCD✳ ❈➙✉ ✽ ✭✶✱✵ ✤✐➸♠✮✳ ●✐↔✐ ❤➺ ♣❤÷ì♥❣ tr➻♥❤ +√ y =1 4−x+2 y + x + y2 y − y + = 2√x + (x, y ∈ R) ❈➙✉ ✾ ✭✶ ✤✐➸♠✮✳ ❈❤♦ x✱ y ❧➔ ❝→❝ sè t❤ü❝ ❞÷ì♥❣ ♣❤➙♥ ❜✐➺t t❤ä❛ ♠➣♥ x2 + 4y2 ≤ 2(xy + 2)✳ ❚➻♠ ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ ❜✐➸✉ t❤ù❝ P = 16 16 + 4+ x4 y (x − y)4 ◆❣✉②➵♥ ❉÷ ❚❤→✐✱ ❚❚❇❉❑❚ ❈❛♦ ❚❤➢♥❣✱ ✶✶ ✣è♥❣ ✣❛✱ ❚P ❍✉➳✱ ❉✣✿ ✵✾✵✺✾✾✽✸✻✾ ✣⑩P ⑩◆ ✣➋ ❚❍■ ❚❍Û ❙➮ ✶✻ ❈➙✉ ✶✳ ❛✮ ❜✮ ❈❤♦ ❤➔♠ sè x3 − 6x2 + 3(4 − m2)x + 6m2 − ✭✶✮✱ tr♦♥❣ ✤â m ❧➔ t❤❛♠ sè✳ ❑❤↔♦ st sỹ t ỗ t số m = m ỗ t❤à ❤➔♠ sè ✭✶✮ ✤↕t ❝ü❝ trà t↕✐ A✱ B s❛♦ ❝❤♦ t❛♠ ❣✐→❝ OAB ✈✉ỉ♥❣ t↕✐ O✳ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ❛✮ ❱ỵ✐ m = t❛ ❝â y = x3 − 6x2 + 9x − 1✳ • ❚➟♣ ①→❝ ✤à♥❤✿ R✳ • ❚❛ ❝â y = 3x2 − 12x + 9, y =0⇔ • • lim y = lim x3 − x→±∞ x→±∞ ❇↔♥❣ ❜✐➳♥ t❤✐➯♥✿ x x = ⇒ y(1) = x = ⇒ y(3) = −1 = ±∞✳ + 2− x x x −∞ + y +∞ − + +∞ y −∞ • • ã ã số ỗ tr (−∞; 1)✱ (3; +∞)✳ ❍➔♠ sè ♥❣❤à❝❤ ❜✐➳♥ tr➯♥ (1; 3) ỗ t số t ỹ t (1; 3) t ỹ t t (3; 1) ỗ t y 3 x O −1 ✶ ❜✮ ❚➟♣ ①→❝ ✤à♥❤✿ R✳ y • ❚❛ ❝â = 3x2 − 12x + 3(4 − m2 )✳ x = − m ⇒ y(2 − m) = + 2m3 x = + m ⇒ y(2 + m) = − 2m3 y =0⇔ • • ❍➔♠ sè ❝â ❤❛✐ ✤✐➸♠ ❝ü❝ trà ⇔ y = ❝â ❤❛✐ ♥❣❤✐➺♠ ♣❤➙♥ ❜✐➺t ⇔ − m = + m ⇔ m = 0✳ ❑❤✐ ✤â ❤❛✐ ✤✐➸♠ ❝ü❝ tr ỗ t số A(2 m; + 2m3)✱ B(2 + m; − 2m3)✳ ❚❛♠ ❣✐→❝ OAB ✈✉æ♥❣ t↕✐ O −→ −→ ⇔OA · OB= ⇔ (2 − m)(2 + m) + (1 + 2m3 )(1 − 2m3 ) = ⇔ −4m6 − m2 + = ⇔ m2 = ⇔ m = ±1 • ❱➟② m = ±1✳ ❈➙✉ ✷✳ ❛✮ ❜✮ ●✐↔✐ ❜➜t ♣❤÷ì♥❣ tr➻♥❤ 52x+1 − 26 · 5x + > (x ∈ R)✳ ❚➻♠ t➟♣ ❤ñ♣ ❝→❝ ✤✐➸♠ M ❜✐➸✉ ❞✐➵♥ sè ♣❤ù❝ z ❜✐➳t |z − 1| = |(1 − i)z|✳ P❤➙♥ t➼❝❤✲▲í✐ t ữỡ tr tữỡ ữỡ ợ à (5x )2 − 26 · 5x + > x < x < −1 ⇔ ⇔ x > x >5 ❜✮ ❱➟② t➟♣ ♥❣❤✐➺♠ ❝õ❛ ❜➜t ♣❤÷ì♥❣ tr➻♥❤ ❧➔ S = (−∞; −1) ∪ [1; +∞) ✣➦t z = x + yi✱ x, y ∈ R✳ ❚❛ ❝â M (x, y) ✈➔ |z − 1| = |(1 − i)z| ⇔ |(x − 1) + yi| = |1 − i| · |x + yi| √ ⇔ (x − 1)2 + y = · x2 + y ⇔ x2 + y − 2x + = 2x2 + 2y ⇔ x2 + y + 2x − = ❱➟② t➟♣ ❤ñ♣ ❝→❝ ✤✐➸♠ M ❜✐➸✉√❞✐➵♥ sè ♣❤ù❝ z ❧➔ ✤÷í♥❣ trá♥ (C) : x2 + y2 + 2x − = ❝â t➙♠ I(−1; 0) ✈➔ ❝â ❜→♥ ❦➼♥❤ R = 2✳ ❈➙✉ ✸✳ ❚➼♥❤ t➼❝❤ ♣❤➙♥ I= 2x · √ √ ex ❞x P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ❚❛ ❝â I= • ex + x2 + + 2x √ ex (2x + 1) + x2 ❞x = + 2x ❚➼♥❤ A✿ A= 2√ e x ❞x + e x ❞x = 2e x 1 ✷ 2 x2 ❞x = A + B + 2x = 2(e − 1) • ❚➼♥❤ B ✿ ✣➦t t = 2x + ⇒ x = t −2 ⇒ ❞x = 12 ❞t✳ ✣ê✐ ❝➟♥✿ B= 1 = (t − 1)2 1 · ❞t = 4t t − 2t + ln t x t t t−2+ = ❞t 1 + ln ❱➟② I = 2e − 23 + 81 ln 5✳ ❚r♦♥❣ ❦❤ỉ♥❣ ❣✐❛♥ ✈ỵ✐ ❤➺ tå❛ ✤ë Oxyz✱ ❝❤♦ ♠➦t ♣❤➥♥❣ (P ) : x − 2y + 2z − = ✈➔ ❤❛✐ ✤÷í♥❣ −3 z x−5 y z+5 t❤➥♥❣ d1 : x −2 = y−3 = ✱ d2 : = = ✳ −5 ❈➙✉ ✹✳ ❛✮ ❜✮ ❱✐➳t ♣❤÷ì♥❣ tr➻♥❤ ♠➦t ♣❤➥♥❣ (Q) q✉❛ ✤÷í♥❣ t❤➥♥❣ d1 ✈➔ ✈✉ỉ♥❣ ❣â❝ ✈ỵ✐ ♠➦t ♣❤➥♥❣ (P )✳ ❚➻♠ ✤✐➸♠ M t❤✉ë❝ d1 ✈➔ ✤✐➸♠ N t❤✉ë❝ d2 s❛♦ ❝❤♦ ữớ t M N s s ợ t (P ) ỗ tớ ữớ t M N t (P ) ♠ët ❦❤♦↔♥❣ ❜➡♥❣ ✷✳ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ❛✮ −→ ✣÷í♥❣ t❤➥♥❣ d1 q✉❛ ✤✐➸♠ A(1; 3; 0)✱ ♥❤➟♥ u = (2; −3; 1) ❧➔♠ ✈❡❝tì ❝❤➾ ♣❤÷ì♥❣✳ ▼➦t ♣❤➥♥❣ (P ) ❝â ✈❡❝tì ♣❤→♣ t✉②➳♥ ❧➔ −→ nP = (1; −2; 2)✳ ❱➻ (Q) ❝❤ù❛ ✤÷í♥❣ t❤➥♥❣ d1 ✈➔ ✈✉ỉ♥❣ ❣â❝ ✈ỵ✐ ♠➦t ♣❤➥♥❣ (P ) ♥➯♥ ✤✐ q✉❛ A ✈➔ ❝â ✈❡❝tì ♣❤→♣ t✉②➳♥ ❧➔ −→ −→ −→ n = nP , u ❜✮ = (4; 3; 1) ❉♦ ✤â ♣❤÷ì♥❣ tr➻♥❤ ♠➦t ♣❤➥♥❣ (Q) : 4x + 3y + z − 13 = 0✳ ❚❛ ❝â M ∈ d1 ⇒ M (2a + 1; −3a + 3; a)✱ N ∈ d2 ⇒ N (6b + 5; 4b; −5b − 5)✱ −→ M N = (6b − 2a + 4; 4b + 3a − 3; −5b − a − 5) • ❚❛ ❝â MN (P ) ⇔ M∈ / (P ) −→ −→ MN · n = (2a + 1) − 2(−3a + 3) + 2a − = (6b − 2a + 4) − 2(4b + 3a − 3) + 2(−5b − a − 5) = a= ⇔ 6b + 5a = ⇔ • ❱➻ M N (P ) ♥➯♥ ❞(M N, (P )) = ❞(M, (P )) = |10a3− 6| ✳ ❉♦ ✤â ❞(M N, (P )) = ⇔ |10a − 6| = ⇔ • • a=0 a= ❱ỵ✐ a = t❛ ❝â b = ♥➯♥ M (1; 3; 0)✱ N (5; 0; −5)✳ ❱ỵ✐ a = 56 t❛ ❝â b = −1 ♥➯♥ M 175 ; − 35 ; 65 ✱ N (−1; −4; 0)✳ ✸ ❱➟② M (1; 3; 0)✱ N (5; 0; −5) ❤♦➦❝ M 17 ;− ; 5 ✱ N (−1; −4; 0)✳ ❍➻♥❤ ❝❤â♣ S.ABC ❝â ✤→② ABC ❧➔ t❛♠ ❣✐→❝ ✈✉æ♥❣ t↕✐ B ✱ BC = SA = a✱ ACB = SAC = 600✱ ♠➦t ♣❤➥♥❣ (SAC) ✈✉ỉ♥❣ ❣â❝ ✈ỵ✐ ♠➦t ✤→②✳ ❚➼♥❤ t❤❡♦ a t❤➸ t➼❝❤ ❦❤è✐ ❝❤â♣ S.ABC ✈➔ ❦❤♦↔♥❣ ❝→❝❤ tø ✤✐➸♠ A ✤➳♥ ♠➦t ♣❤➥♥❣ (SBC)✳ ❈➙✉ ✺✳ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ S T H C A K B ●å✐ H ❧➔ ❤➻♥❤ ❝❤✐➳✉ ✈✉æ♥❣ ❣â❝ ❝↔✉ S tr➯♥ AC ✱ K ❧➔ ❤➻♥❤ ❝❤✐➳✉ ✈✉æ♥❣ ❣â❝ ❝õ❛ H tr➯♥ BC ✈➔ T ❧➔ ❤➻♥❤ ❝❤✐➳✉ ✈✉æ♥❣ ❣â❝ ❝õ❛ H tr➯♥ SK ✳ • ❚❛ ❝â (SAC) ⊥ (ABC) (SAC) ∪ (ABC) = AC SH ∈ (SAC) SH ⊥ AC √ AB = AC tan ACB = a , AC = ❉♦ ✤â VS.ABC • AB ⇒ SH ⊥ (ABC) = 2a , AH = SA cos SAC = sin ACB √ √ a2 a SH = SA sin SAC = , SABC = BC · BA = 2 a3 = SH · SABC = ✳ ❚❛ ❝â BC ⊥ HK ✈➔ BC (SBC)✳ ❚❤➔♥❤ t❤û AC ❞(A, (SBC)) = HC · ❞(H, (SAC)) = · ❞(H, (SAC)) ⊥ SH ♥➯♥ BC ⊥ (SHK)✳ ❙✉② r❛ BC ⊥ HT ✳ ▼➔ SK ⊥ HT ❞(H, (SBC)) = HT √ 3a • HK = CH sin ACB = ✳ HT ❧➔ ✤÷í♥❣ ❝❛♦ ❝õ❛ t❛♠ ❣✐→❝ ✈✉ỉ♥❣ SHK ♥➯♥ √ SH · HK SH · HK 3a 39 HT = =√ = SK 26 SH + HK ❱➟② a 3a , HC = AC−AH = , 2 √ 2a 39 ❞(A, (SBC)) = 13 ✳ ✹ ♥➯♥ HT ⊥ ❈➙✉ ✻✳ ❛✮ ❜✮ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ sin 3x + cos 2x = sin x, x ∈ (−π; π)✳ ▼ët ❧ỵ♣ ❝â ✸✵ ❤å❝ s✐♥❤ tr♦♥❣ ✤â ❝â ✸ ❤å❝ s✐♥❤ ❧➔ ❝→♥ ❜ë ❧ỵ♣✳ ❈❤å♥ ♥❣➝✉ ♥❤✐➯♥ ✸ ❤å❝ s✐♥❤ tø ❧ỵ♣ ✤➸ ❧➔♠ trü❝ ♥❤➟t ❧ỵ♣ ❤å❝✳ ❚➼♥❤ ①→❝ s✉➜t ✤➸ tr♦♥❣ ✸ ❤å❝ s✐♥❤ ✤÷đ❝ ❝❤å♥ ❝â ➼t ♥❤➜t ♠ët ợ P tớ õ ữỡ tr tữỡ ữỡ ợ (sin 3x sin x) + cos 2x = ⇔ cos 2x sin x + cos 2x = cos 2x = ⇔ cos 2x · (2 sin x + 3) = ⇔ sin x = − (❧♦↕✐) π π π ⇔2x = + kπ, k ∈ Z ⇔ x = + k , k ∈ Z ✳ ▼➔ k ∈ Z − ; 2 3π π π 3π − ;− ; ; ✳ 4 4 ❚❛ ❝â x ∈ (−π; π) ⇔ k ∈ ♣❤÷ì♥❣ tr➻♥❤ ❧➔ S = ❜✮ ♥➯♥ k ∈ {−2; −1; 0; 1}✳ ❱➟② t➟♣ ♥❣❤✐➺♠ ❝õ❛ ●å✐ Ω ❧➔ ❦❤æ♥❣ ❣✐❛♥ ♠➝✉✱ A ❧➔ ❜✐➳♥ ❝è ✧tr♦♥❣ ✸ ❤å❝ s✐♥❤ ✤÷đ❝ ❝❤å♥ ❝â ➼t ♥❤➜t ♠ët ❝→♥ ❜ë ❧ỵ♣✧✳ ❚➼♥❤ |Ω|✿ |Ω| ❝❤➼♥❤ ❧➔ sè ❝→❝❤ ❝❤å♥ ✸ ❤å❝ s✐♥❤ tø ♠ët ❧ỵ♣ ❝â ✸✵ ❤å❝ s✐♥❤ ♥➯♥ |Ω| = C30 = 4060 ❚r♦♥❣ sè C303 ❝→❝❤ ❝❤å♥ ✸ ❤å❝ s✐♥❤ tũ ỵ õ C273 ổ õ ❧ỵ♣ ♥➔♦✳ ❉♦ ✤â sè ❝→❝❤ ❝❤å♥ ✸ ❤å❝ s✐♥❤ ❝â ➼t ♥❤➜t ♠ët ❝→♥ ❜ë ❧ỵ♣ ❧➔ ❚➼♥❤ |ΩA |✿ 3 |ΩA | = C30 − C27 = 1135 ❱➟② ①→❝ s✉➜t ❝õ❛ ❜✐➳♥ ❝è A ❧➔ P (A) = |ΩA | 227 = |Ω| 812 ❚r♦♥❣ ♠➦t ♣❤➥♥❣ ✈ỵ✐ ❤➺ tå❛ ✤ë Oxy✱ ❝❤♦ ❤➻♥❤ ✈✉ỉ♥❣ ABCD ❝â ✤✐➸♠ B t❤✉ë❝ ✤÷í♥❣ t❤➥♥❣ d : x + 3y − = 0✱ ✤✐➸♠ E t❤✉ë❝ t✐❛ ✤è✐ ❝õ❛ t✐❛ BA s❛♦ ❝❤♦ BA = 2BE ✱ ✤✐➸♠ H(7; 3) ❧➔ ❤➻♥❤ ❝❤✐➳✉ ✈✉æ♥❣ ❣â❝ ❝õ❛ A tr➯♥ CE ✱ ❤❛✐ ✤÷í♥❣ t❤➥♥❣ DH ✈➔ AB ❝➢t ♥❤❛✉ t↕✐ ✤✐➸♠ F (16; 0)✳ ❚➻♠ tå❛ ✤ë ❝→❝ ✤➾♥❤ ❝õ❛ ❤➻♥❤ ✈✉æ♥❣ ABCD✳ ❈➙✉ ✼✳ D C H A B E ✺ F P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ❈→❝ tù ❣✐→❝ ABHC ✈➔ ADCH ❝ị♥❣ ♥ë✐ t✐➳♣ ✤÷í♥❣ trá♥ ✤÷í♥❣ ❦➼♥❤ AC ♥➯♥ tù ❣✐→❝ DCHB ❝ơ♥❣ ♥ë✐ t✐➳♣ ✤÷í♥❣ trá♥ ✤÷í♥❣ ❦➼♥❤ AC ✳ ❉♦ ✤â DHC = DCB = 900 , BHE = BCD = 450 • ❚❛ ❝â DF : x + 3y − 16 = 0✱ BH : 3x − y − 18 = 0✱ B ❧➔ ❣✐❛♦ ✤✐➸♠ ❝õ❛ DF ✈ỵ✐ BH ♥➯♥ B(6; 0)✳ ❧➔ ♣❤➙♥ ❣✐→❝ ❣â❝ BHF ♥➯♥ t❛ ❝â HE −→ BH 1 −→ BE = = ⇒BE= BF ⇒ E EF FH 17 ;0 • EA= EB⇒ A(1; 0)✳ −→ −→ • AD : x = 1✱ D ❈➙✉ ✽✳ ❧➔ ❣✐❛♦ ✤✐➸♠ ❝õ❛ AD ✈ỵ✐ DF ♥➯♥ D(1; 5) ⇒ C(6; 5)✳ y +√ =1 4−x+2 y+ x+ √ y −y+4=2 x+1 ●✐↔✐ ❤➺ ♣❤÷ì♥❣ tr➻♥❤ (1) y2 (x, y ∈ R) (2) P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ✣✐➲✉ ❦✐➺♥✿ −1 ≤ x ≤ x + y2 ≥ y + x + y = ✳ ❑❤✐ ✤â ❤➺ trð t❤➔♥❤ ❚r÷í♥❣ ❤đ♣ ✶✿ x = y =1 y+ y −y+2=0 y2 + (✈æ ♥❣❤✐➺♠) ✳ ❚❛ ❝â ❚r÷í♥❣ ❤đ♣ ✷✿ x = x + y2 − y (1) ⇔ + y 2− √ 4−x x x √ x+y =y 4−x+x ⇔2 =1 √ ⇒ 4x + 4y = 4y − xy + x2 + 2xy − x √ √ ⇒ y + (4 − x) − 2y − x = ⇒ (y − − x)2 = ⇒y= ❚❤❛② x = − y2 ✈➔♦ ✭✷✮ t❛ ❝â ✣✐➲✉ ❦✐➺♥ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ✭✸✮✿ √ 4−x⇒ y≥0 x = − y2 y − y + − − y = (3) √ ≤ y ≤ 5✳ ❚❛ ❝â (3) ⇔(y − y) + 2 − ⇔(y − 1)(y + y) + ⇔(y − 1) y + y + ❱➻ y2 + y + 2(y + 1) 2+ 5−y ❝õ❛ ❤➺ ❧➔ (x, y) = (3; 1)✳ >0 ✈ỵ✐ ♠å✐ y ∈ [0; √ 5] − y2 = 2(y − 1)(y + 1) 2+ − y2 2(y + 1) 2+ − y2 =0 = (4) ♥➯♥ ✭✹✮⇔ y = ⇒ x = ✭t❤ä❛ ♠➣♥✮✳ ❱➟② ♥❣❤✐➺♠ ✻ ❈❤♦ x✱ y ❧➔ ❝→❝ sè t❤ü❝ ❞÷ì♥❣ ♣❤➙♥ ❜✐➺t t❤ä❛ ♠➣♥ x2 + 4y2 ≤ 2(xy + 2)✳ ❚➻♠ ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ ❜✐➸✉ t❤ù❝ 16 16 ❈➙✉ ✾✳ P = x4 + y4 + (x − y)4 P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ❚❛ ❝â 4xy ≤ x2 + 4y ≤ 2(xy + 2) ⇒ xy ≤ ❉♦ ✤â 4P ≥ x2 y = 16 = 16 16 16 + 4+ x4 y (x − y)4 y2 5x2 y x2 + + y2 x2 (x2 + y − 2xy)2 x y + y x −2 + x y + −2 y x ✣➦t t = xy + xy ✱ t > 2✳ ❚❛ ❝â 4P ≥ 16t2 + − 32 (t − 2)2 ❚❤❡♦ ❜➜t ✤➥♥❣ t❤ù❝ ❈❛✉❝❤②✱ t❛ ❝â 1 + 8(t − 2) + 8(t − 2) ≥ 3 · 8(t − 2) · 8(t − 2) = 12 ⇒ ≥ 44 − 16t 2 (t − 2) (t − 2) (t − 2)2 ❚ø ✤â ❝â 4P ≥ 16t2 − 80t + 188 = (2t − 5)2 + 78 ≥ 88 ⇒ P ≥ 22 ❑❤✐ x = ✈➔ y = t❤➻ P = 22 ♥➯♥ ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ P ❜➡♥❣ ✷✷✳ ✼ ... ❚➻♠ ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ ❜✐➸✉ t❤ù❝ 16 16 ❈➙✉ ✾✳ P = x4 + y4 + (x − y)4 P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ❚❛ ❝â 4xy ≤ x2 + 4y ≤ 2(xy + 2) ⇒ xy ≤ ❉♦ ✤â 4P ≥ x2 y = 16 = 16 16 16 + 4+ x4 y (x − y)4 y2 5x2 y x2... 2✳ ❚❛ ❝â 4P ≥ 16t2 + − 32 (t − 2)2 ❚❤❡♦ ❜➜t ✤➥♥❣ t❤ù❝ ❈❛✉❝❤②✱ t❛ ❝â 1 + 8(t − 2) + 8(t − 2) ≥ 3 · 8(t − 2) · 8(t − 2) = 12 ⇒ ≥ 44 − 16t 2 (t − 2) (t − 2) (t − 2)2 ❚ø ✤â ❝â 4P ≥ 16t2 − 80t + 188... ❛✮ ❜✮ ❈❤♦ ❤➔♠ sè x3 − 6x2 + 3(4 − m2)x + 6m2 − ✭✶✮✱ tr♦♥❣ ✤â m ❧➔ t số st sỹ t ỗ t❤à ❝õ❛ ❤➔♠ sè ✭✶✮ ❦❤✐ m = 1✳ ❚➻♠ m ỗ t số t ỹ tr t↕✐ A✱ B s❛♦ ❝❤♦ t❛♠ ❣✐→❝ OAB ✈✉æ♥❣ t↕✐ O✳ P❤➙♥ t➼❝❤✲▲í✐