Để chuẩn bị tốt những kiến thức cơ bản để cho kì thi Đại học sắp tới mời các bạn cùng tham khảo Đề thi thử Đại học năm 2015 môn Toán - Đề số 17. 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 y = x − 8x ✭✶✮✳ ❛✮ st sỹ t ỗ t (C) ❝õ❛ ❤➔♠ sè ✭✶✮✳ ❜✮ ❱✐➳t ♣❤÷ì♥❣ tr➻♥❤ t✐➳♣ t✉②➳♥ ợ ỗ t (C) t õ a ♠➔ f ❈➙✉ ✷ ✭✶✱✵ ✤✐➸♠✮✳ ❛✮ ❈❤♦ x ∈ π4 ; π2 t❤ä❛ ♠➣♥ sin x + cos x = 75 ✳ ❚➼♥❤ A = sin x − cos x✳ (a) = −4✳ ❜✮ ❚➻♠ sè ♣❤ù❝ z ❜✐➳t z − +z i = − 3i✳ ❈➙✉ ✸ ✭✵✱✺ ✤✐➸♠✮✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ log (x + 1) + log (2x + 3) = (x ∈ R)✳ ❈➙✉ ✹ ✭✶✱✵ ✤✐➸♠✮✳ ❚➻♠ m ✤➸ ♣❤÷ì♥❣ tr➻♥❤ s❛✉ ❝â ♥❣❤✐➺♠ m √ √ √ √ − x − + x + 12 − x2 = 2(8x + m) + 15 (x ∈ R) ❈➙✉ ✺ ✭✶✱✵ ✤✐➸♠✮✳ ❚➼♥❤ t➼❝❤ ♣❤➙♥ I= x + ln x ❞x (1 + x)2 ❈➙✉ ✻ ✭✶✱✵ ✤✐➸♠✮✳ ❚r♦♥❣ ♠➦t ♣❤➥♥❣ ✈ỵ✐ ❤➺ tå❛ ✤ë Oxy✱ ❝❤♦ t❛♠ ❣✐→❝ ABC ❝â t➙♠ ✤÷í♥❣ trá♥ ♥❣♦↕✐ t✐➳♣ ❧➔ I(3; 5)✱ ✤÷í♥❣ ♣❤➙♥ ❣✐→❝ tr♦♥❣ ❣â❝ A ❧➔ d : 3x + y − = 0✱ ❝❤➙♥ ✤÷í♥❣ ✈✉ỉ♥❣ ❣â❝ ❤↕ tø A ❧➔ D(0; 2)✳ ❚➻♠ tå❛ ✤ë ❝→❝ ✤➾♥❤ A✱ B ✱ C ✳ ❈➙✉ ✼ ✭✶✱✵ ✤✐➸♠✮✳ ❍➻♥❤ ❝❤â♣ S.ABCD ❝â ✤→② ❧➔ ❤➻♥❤ ✈✉æ♥❣ ❝↕♥❤ a✱ ❤❛✐ ♠➦t ♣❤➥♥❣ (SAB) ✈➔ (SAD) ❝ị♥❣ ✈✉ỉ♥❣ ❣â❝ ✈ỵ✐ ♠➦t ✤→②✱ ♠➦t ♣❤➥♥❣ (SBC) t↕♦ ✈ỵ✐ ♠➦t ✤→② ♠ët ❣â❝ 600✳ ❚➼♥❤ t❤❡♦ a t❤➸ t➼❝❤ ❦❤è✐ ❝❤â♣ S.ABCD ✈➔ ❦❤♦↔♥❣ ❝→❝❤ ❣✐ú❛ ❤❛✐ ✤÷í♥❣ t❤➥♥❣ SC ✱ BD✳ ❈➙✉ ✽ ✭✶✱✵ ✤✐➸♠✮✳ ❚r♦♥❣ ❦❤ỉ♥❣ ❣✐❛♥ ✈ỵ✐ ❤➺ tå❛ ✤ë Oxyz✱ ❝❤♦ ♠➦t ♣❤➥♥❣ (P ) : x + y + z + = ✈➔ ✤✐➸♠ J(1; −2; −4)✳ ❛✮ ❱✐➳t ♣❤÷ì♥❣ tr➻♥❤ ✤÷í♥❣ t❤➥♥❣ d q✉❛ J ✈➔ ✈✉ỉ♥❣ ❣â❝ ✈ỵ✐ ♠➦t ♣❤➥♥❣ (P )✳ ❜✮ ❱✐➳t ♣❤÷ì♥❣ tr➻♥❤ ♠➦t ❝➛✉ (S) ❝â ❜→♥ ❦➼♥❤ R = ✈➔ ❝➢t ♠➦t ♣❤➥♥❣ (P ) t❤❡♦ ❣✐❛♦ t✉②➳♥ ❧➔ ✤÷í♥❣ trá♥ √ ❝â t➙♠ J ✈➔ ❝â ❜→♥ ❦➼♥❤ ❜➡♥❣ r = 13✳ ❈➙✉ ✾ ✭✵✱✺ ✤✐➸♠✮✳ ❈â ❤❛✐ ❧æ ❤➔♥❣✱ ❧æ ❤➔♥❣ t❤ù ♥❤➜t ❝â ✶✷ s↔♥ ♣❤➞♠ ❦❤→❝ ♥❤❛✉ tr♦♥❣ ✤â ❝â ✷ ♣❤➳ ♣❤➞♠✱ ❧æ ❤➔♥❣ t❤ù ❤❛✐ ❝â ✽ s↔♥ ♣❤➞♠ ❦❤→❝ ♥❤❛✉ tr♦♥❣ ✤â ❝â ✶ ♣❤➳ ♣❤➞♠✳ ▲➜② ♥❣➝✉ ♥❤✐➯♥ tø ♠é✐ ❧æ ❤➔♥❣ ✷ s↔♥ ♣❤➞♠✳ ❚➼♥❤ ①→❝ s✉➜t ✤➸ tr♦♥❣ ✹ s↔♥ ♣❤➞♠ ❧➜② r❛ ❝â ❦❤æ♥❣ q✉→ ✶ ♣❤➳ ♣❤➞♠✳ ❈➙✉ ✶✵ ✭✶ ✤✐➸♠✮✳ ❈❤♦ x✱ y✱ z ❧➔ ❝→❝ sè t❤ü❝ t❤ä❛ ♠➣♥ x, y, z ≥ −2 ✈➔ x3 + y3 + z3 ≥ x2 + y2 + z2✳ ❈❤ù♥❣ ♠✐♥❤ x5 + y + z ≥ x2 + y + z ◆❣✉②➵♥ ❉÷ ❚❤→✐✱ ❚❚❇❉❑❚ ❈❛♦ ❚❤➢♥❣✱ ✶✶ ✣è♥❣ ✣❛✱ ❚P ❍✉➳✱ ❉✣✿ ✵✾✵✺✾✾✽✸✻✾ ✣⑩P ⑩◆ ✣➋ ❚❍■ ❚❍Û ❙➮ ✶✼ ❈➙✉ ✶✳ ❈❤♦ ❤➔♠ sè y = x4 − 8x2 st sỹ t ỗ t❤à (C) ❝õ❛ ❤➔♠ sè ✭✶✮✳ ❜✮ ❱✐➳t ♣❤÷ì♥❣ tr➻♥❤ t t ợ ỗ t (C) t õ ✤ë a ♠➔ y (a) = −4✳ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ❛✮ • • • • ❚➟♣ ①→❝ ✤à♥❤✿ R✳ ❚❛ ❝â y = 4x3 − 16x, y = ⇔ x ∈ {−2; 0; 2} lim y = lim x4 − x→±∞ x→±∞ ❇↔♥❣ ❜✐➳♥ t❤✐➯♥✿ x −∞ = +∞✳ −2 − y x2 + − f +∞ + +∞ y −16 ã ã ã ã 16 số ỗ tr ❝→❝ ❦❤♦↔♥❣ (−2; 0) ✈➔ (2; +∞)✳ ❍➔♠ sè ♥❣❤à❝❤ tr (; 2) (0; 2) ỗ t❤à ❤➔♠ sè ✤↕t ❝ü❝ ✤↕✐ t↕✐ (0; 0) ✈➔ t ỹ t t (2; 16) (2; 16) ỗ t y −2 O x −16 ✶ +∞ ❜✮ ❚❛ ❝â y • = 12x2 − 16 ❚❛ ❝â ♥➯♥ y (a) = 12a2 − 16✳ ●å✐ A ❧➔ ✤✐➸♠ t❤✉ë❝ (C) ✈➔ ❝â ❤♦➔♥❤ ✤ë a✳ y (a) = −4 ⇔ a2 = ⇔ • a = −1 a=1 ❱ỵ✐ a = −1✱ t❛ ❝â A(−1; −7)✱ y (−1) = 12 ♥➯♥ t✐➳♣ t✉②➳♥ ✈ỵ✐ (C) t↕✐ A ❧➔ d1 : y = y (−1)(x + 1) − = 12x + ã ợ a = t❛ ❝â A(1; −7)✱ y (−1) = −12 ♥➯♥ t✐➳♣ t✉②➳♥ ✈ỵ✐ (C) t↕✐ A ❧➔ d2 : y = y (−1)(x − 1) − = −12x + ❈➙✉ ✷✳ ❛✮ ❈❤♦ x ∈ π π ; t❤ä❛ ♠➣♥ sin x + cos x = 75 ✳ ❚➼♥❤ A = sin3 x − cos3 x✳ ❜✮ ❚➻♠ sè ♣❤ù❝ z ❜✐➳t z − 7+i = − 3i z ✭✶✮ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ❛✮ ❚❛ ❝â (sin x − cos x)2 = sin2 x + cos2 x − sin x cos x = − (sin x + cos x)2 = ❱➻ x ∈ π π ; 25 ♥➯♥ sin x − cos x > 0✳ ❉♦ ✤â sin x − cos x = 15 ✳ ❚ø ✤â ❝â sin x = , cos x = 5 37 ❱➟② A = 125 ✳ ❜✮ ✣✐➲✉ ❦✐➺♥✿ z = 0✳ ✣➦t z = x + yi✱ x, y ∈ R✳ ❚❛ ❝â (1) ⇔(x + yi)(x − yi) − (7 + i) = (2 − 3i)(x + yi) ⇔ x2 + y − − i = 2x + 3y + (2y − 3x)i 13x2 − 32x − 21 = 2 x + y − = 2x + 3y ⇔ ⇔ 3x − y = 2y − 3x = −1 17 ⇔(x, y) ∈ (3; 4); − ; − 13 13 17 ❱➟② z = + 4i ❤♦➦❝ z = − 137 − 13 i✳ ❈➙✉ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ log3 (x + 1)2 + log√3 (2x + 3) = (x ∈ R)✳ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ✣✐➲✉ ❦✐➺♥✿ x>− x = −1 P❤÷ì♥❣ tr➻♥❤ trữỡ ữỡ ợ log3 |x + 1| + log3 (2x + 3) = ⇔ log3 |x + 1|(2x + 3) = ⇔ |x + 1|(2x + 3) = (x + 1)(2x + 3) = −3 2x2 + 5x + = (✈æ ♥❣❤✐➺♠) ⇔ ⇔ ⇔ x ∈ 0; − (x + 1)(2x + 3) = 2x + 5x = ❱➟② ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❧➔ x = 0✳ ✷ ❈➙✉ ✹✳ ❚➻♠ m ✤➸ ♣❤÷ì♥❣ tr➻♥❤ s❛✉ ❝â ♥❣❤✐➺♠ m √ √ − x − + x + 12 − x2 = 2(8x + m) + 15 (x ∈ R) P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ✣✐➲✉ ❦✐➺♥✿ x ∈ D = [−1; 1]✳ √ √ ✣➦t t = − x − + x✱ t❛ ❝â t2 = − x + + 9x − − x2 ⇒ 8x − − x2 = t2 − 10, t =− √ − √ < 0, ∀x ∈ (−1; 1) 1−x 1+x ❇↔♥❣ ❜✐➳♥ t❤✐➯♥✿ x −1 − t √ t ❚ø ❜↔♥❣ ❜✐➳♥ t❤✐➯♥ t❛ ❝â t ∈ √ √ −3 2; √ −3 ✳ ❚❛ â ♣❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ trð t❤➔♥❤ m(t − 2) = 2(t2 − 10) + 15 ⇔ m = ❳➨t ❤➔♠ sè f (t) = 2tt −−25 ✱ t ∈ √ √ −3 2; 2t2 − (1) t−2 ✳ ❚❛ ❝â (1) ⇔ m = f (t) (2) P❤÷ì♥❣ tr➻♥❤ ✤➣ ❝❤♦ ❝â ♥❣❤✐➺♠ ⇔ P❤÷ì♥❣ tr➻♥❤ ✭✷✮ ❝â ♥❣❤✐➺♠ t ∈ ❚❛ ❝â √ 4− 2t2 − 8t + t = 2√ f (t) = =0⇔ 4+ (t − 2)2 t= ❇↔♥❣ ❜✐➳♥ t❤✐➯♥✿ x √ 4− √ −3 + t √ √ −3 2; (3) (❧♦↕✐) √ − √ 8−2 t ❚ø ❜↔♥❣ ❜✐➳♥ t❤✐➯♥ t❛ ❝â √ 62 − 93 14 √ √ 62 − 93 (3) ⇔ m ∈ ;8 − 14 ❈➙✉ ✺✳ ❚➼♥❤ t➼❝❤ ♣❤➙♥ I= √ 2+ 2 ✳ x + ln x ❞x (1 + x)2 ✸ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ✣➦t✿ u = x + ln x ✈➔ ❞v = (x +1 1)2 ❞x✱ t❛ ❝â ❞u = x +x ❞x ✈➔ v = − x +1 ✳ ❉♦ ❞â x + ln x I=− x+1 + 1 ❞x x 1 = − − ln + + ln x 3 2 = ln − ❈➙✉ ✻✳ ❚r♦♥❣ ♠➦t ♣❤➥♥❣ ✈ỵ✐ ❤➺ tå❛ ✤ë Oxy ✱ ❝❤♦ t❛♠ ❣✐→❝ ABC ❝â t➙♠ ✤÷í♥❣ trá♥ ♥❣♦↕✐ t✐➳♣ ❧➔ ✤÷í♥❣ ♣❤➙♥ ❣✐→❝ tr♦♥❣ ❣â❝ A ❧➔ d : 3x + y − = 0✱ ❝❤➙♥ ✤÷í♥❣ ✈✉ỉ♥❣ ❣â❝ ❤↕ tø A ❧➔ D(0; 2)✳ ❚➻♠ tå❛ ✤ë ❝→❝ ✤➾♥❤ A✱ B ✱ C ✳ I(3; 5)✱ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ A I H F M B C D E ●å✐ E ❧➔ ❣✐❛♦ tự d ợ ữớ trỏ (S) t✐➳♣ t❛♠ ❣✐→❝ ABC ✳ • • ❚❛♠ ❣✐→❝ AIE ❝➙♥ t↕✐ I ♥➯♥ IAE = AEI ✳ ❚❛ ❝â EB = EC ✈➔ IB = IC ♥➯♥ IE ❧➔ ✤÷í♥❣ tr✉♥❣ trü❝ ❝õ❛ BC ✳ ❉♦ ✤â IE ⊥ BC ♠➔ AD ⊥ BC ♥➯♥ IE AD✳ ❙✉② r❛ DAE = AEI ✳ ❚❤➔♥❤ t❤û DAE = IAE ✱ ♥❣❤➽❛ ❧➔ d ❧➔ ♣❤➙♥ ❣✐→❝ ❣â❝ DAI ✳ ●å✐ F ố ự ợ I q d ữớ t❤➥♥❣ IF q✉❛ I ✈➔ ✈✉ỉ♥❣ ❣â❝ ✈ỵ✐ d ♥➯♥ ❝â ♣❤÷ì♥❣ tr➻♥❤ IF : x − 3y + 12 = H ❧➔ ❣✐❛♦ ✤✐➸♠ ❝õ❛ IF ✈ỵ✐ d t❤➻ H ; 2 ✳ H ❧➔ tr✉♥❣ ✤✐➸♠ IF ♥➯♥ F (0; 4)✳ ✣÷í♥❣ t❤➥♥❣ AD : x = 0✳ A ❧➔ ❣✐❛♦ ✤✐➸♠ ❝õ❛ AD ợ d A(0; 9) ã õ BC : y = 2✱ (S) : (x − 3)2 + (y − 5)2 = 25✳ B ✈➔ C ❧➔ ❣✐❛♦ BC ợ ữớ trỏ (S) tồ ❝õ❛ B ✈➔ C t❤ä❛ ♠➣♥ ❤➺ y=2 (x − 3)2 + (y − 5)2 = 25 ⇔ (x, y) ∈ {(−1; 2); (7; 2)} ❉♦ ✤â B(−1; 2)✱ C(7; 2) ❤♦➦❝ B(7, 2)✱ C(−1; 2)✳ ❱➟② A(0; 9)✱ B(−1; 2)✱ C(7; 2) ❤♦➦❝ A(0; 9)✱ B(7, 2)✱ C(−1; 2)✳ ❈➙✉ ✼✳ ❍➻♥❤ ❝❤â♣ S.ABCD ❝â ✤→② ❧➔ ❤➻♥❤ ✈✉ỉ♥❣ ❝↕♥❤ a✱ ❤❛✐ ♠➦t ♣❤➥♥❣ (SAB) ✈➔ (SAD) ❝ị♥❣ ✈✉ỉ♥❣ ❣â❝ ✈ỵ✐ ♠➦t ✤→②✱ ♠➦t ♣❤➥♥❣ (SBC) t↕♦ ✈ỵ✐ ♠➦t ✤→② ♠ët ❣â❝ 600✳ ❚➼♥❤ t❤❡♦ a t❤➸ t➼❝❤ ❦❤è✐ ❝❤â♣ S.ABCD ✈➔ ❦❤♦↔♥❣ ❝→❝❤ ❣✐ú❛ ❤❛✐ ✤÷í♥❣ t❤➥♥❣ SC ✱ BD✳ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ S A D E I B C ●å✐ I ❧➔ t➙♠ ❤➻♥❤ ✈✉æ♥❣ ABCD✱ E ❧➔ ❤➻♥❤ ❝❤✐➳✉ ✈✉æ♥❣ ❣â❝ ❝õ❛ I tr➯♥ SC ✳ • ❚❛ ❝â (SAB) ⊥ (ABCD) (SAD) ⊥ (ABCD) (SAB) ∩ (SAD) = SA ⇒ SA ⊥ (ABCD) ❙✉② r❛ BC ⊥ SA✱ ♠➔ BC ⊥ BA ♥➯♥ BC ⊥ (SAB)✳ (SBC) ∩ (ABCD) = BC BC ⊥ (SAB) (SBC) ∩ (SAB) = SB (SAB) ∩ (ABCD) = AB • ⇒ (SBC), (ABCD) = SBA = 600 √ ❚❛ ❝â SA = AB tan SBA = a 3✱ SABCD = a2✳ ❱➟② VS.ABCD √ a3 = SA · SABCD = 3 ✺ • ❚❛ ❝â BD ⊥ AC ✈➔ BD ⊥ SA ♥➯♥ BD ⊥ (SAC) ⇒ BD ⊥ IE ✳ ▼➔ IE ⊥ SC ♥➯♥ ❞(SC, BD) = IE √ √ √ √ a • AC = a 2✱ CS = CA2 + AS = a 5✱ CI = AC = ✳ 2 ❱➟② ❚❛ ❝â √ IE CI SA · CI a 30 ∆CEI ∼ ∆CAS ⇒ = ⇒ IE = = SA SC CS 10 √ a 30 ❞(SC, BD) = 10 ✳ ❈➙✉ ✽✳ ❚r♦♥❣ ❦❤ỉ♥❣ ❣✐❛♥ ✈ỵ✐ ❤➺ tå❛ ✤ë J(1; −2; −4)✳ Oxyz ✱ ❝❤♦ ♠➦t ♣❤➥♥❣ (P ) : x + y + z + = ✈➔ ✤✐➸♠ ❛✮ ❱✐➳t ♣❤÷ì♥❣ tr➻♥❤ ✤÷í♥❣ t❤➥♥❣ d q✉❛ J ✈➔ ✈✉ỉ♥❣ ❣â❝ ợ t (P ) t ữỡ tr t ❝➛✉ ❝â ❜→♥ ❦➼♥❤ R√= ✈➔ ❝➢t ♠➦t ♣❤➥♥❣ (P ) t❤❡♦ ❣✐❛♦ t✉②➳♥ ❧➔ ✤÷í♥❣ trá♥ ❝â t➙♠ J ✈➔ ❝â ❜→♥ ❦➼♥❤ ❜➡♥❣ r = 13✳ (S) P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ❛✮ ❱❡❝tì ♣❤→♣ t✉②➳♥ ❝õ❛ ♠➦t ♣❤➥♥❣ (P ) ❧➔ ♥➯♥ ♥❤➟♥ −→ n n = (1; 1; 1) ữớ t d q J ổ õ ợ (P ) ❧➔♠ ✈❡❝tì ❝❤➾ ♣❤÷ì♥❣✱ ❞♦ ✤â ❝â ♣❤÷ì♥❣ tr➻♥❤ −→ x = + t d : y = −2 + t z = −4 + t ❜✮ ●å✐ I ❧➔ t➙♠ ❝õ❛ ♠➦t ❝➛✉ (S)✳ ❱➻ (S) ❝➢t ♠➦t ♣❤➥♥❣ (P ) t❤❡♦ ❣✐❛♦ t✉②➳♥ ❧➔ ✤÷í♥❣ trá♥ ❝â t➙♠ J ♥➯♥ JI ⊥ (P ) ⇒ IJ = ❞(I, (P )) ✈➔ I ∈ d ⇒ I(a + 1; a − 2; a − 4)✳ ❚❛ ❝â ❞(I, (P )) = R2 − r2 ⇔ JI = √ ⇔3t = t {1; 1} ã ã ợ t = −1 t❛ ❝â I(0; −3; −5) ✈➔ (S) : x2 + (y + 3)2 + (z + 5)2 = 16✳ ❱ỵ✐ t = t❛ ❝â I(2; −1; −3) ✈➔ (S) : (x − 2)2 + (y + 1)2 + (z + 3)2 = 16✳ ❈➙✉ ✾✳ ❈â ❤❛✐ ❧æ ❤➔♥❣✱ ❧æ ❤➔♥❣ t❤ù ♥❤➜t ❝â ✶✷ s↔♥ ♣❤➞♠ ❦❤→❝ ♥❤❛✉ tr♦♥❣ ✤â ❝â ✷ ♣❤➳ ♣❤➞♠✱ ❧æ ❤➔♥❣ t❤ù ❤❛✐ ❝â ✽ s↔♥ ♣❤➞♠ ❦❤→❝ ♥❤❛✉ tr♦♥❣ ✤â ❝â ✶ ♣❤➳ ♣❤➞♠✳ ▲➜② ♥❣➝✉ ♥❤✐➯♥ tø ♠é✐ ❧æ ❤➔♥❣ ✷ s↔♥ ♣❤➞♠✳ ❚➼♥❤ ①→❝ s✉➜t ✤➸ tr♦♥❣ ✹ s↔♥ ♣❤➞♠ ❧➜② r❛ ❝â ❦❤æ♥❣ q✉→ ✶ ♣❤➳ ♣❤➞♠✳ ❑❤æ♥❣ ❣✐❛♥ ♠➝✉ Ω ❝â sè ♣❤➛♥ tû ❧➔ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ |Ω| = C12 · C82 = 1848 ●å✐ A ❧➔ ❜✐➳♥ ❝è ✧tr♦♥❣ ✹ s↔♥ ♣❤➞♠ ❧➜② r❛ ❝â ❦❤æ♥❣ q✉→ ✶ ♣❤➳ ♣❤➞♠✧✳ ❈→❝ ❦❤↔ ♥➠♥❣ t ủ ố A ã ữủ ❝↔ ✹ ❝❤➼♥❤ ♣❤➞♠✿ ❑❤✐ ✤â sè ❝→❝❤ ❝❤å♥ ❧➔ C10 à C72 = 945 ã ữủ ✶ ♣❤➳ ♣❤➞♠ tø ❧æ ❤➔♥❣ t❤ù ♥❤➜t✿ ❑❤✐ ✤â sè ❝→❝❤ ❝❤å♥ ❧➔ C10 · C21 · C72 = 420 ã ữủ tứ ổ ❤➔♥❣ t❤ù ❤❛✐✿ ❑❤✐ ✤â sè ❝→❝❤ ❝❤å♥ ❧➔ C10 · C71 · C11 = 315 ❉♦ ✤â |ΩA| = 945 + 420 + 315 = 1680✳ ❱➟② ①→❝ s✉➜t ❝õ❛ ❜✐➳♥ ❝è A ❧➔ P (A) = |ΩA | 10 = |Ω| 11 ❈➙✉ ✶✵✳ ❈❤♦ x✱ y✱ z ❧➔ ❝→❝ sè t❤ü❝ t❤ä❛ ♠➣♥ x, y, z ≥ −2 ✈➔ x3 + y3 + z ≥ x2 + y2 + z ✳ ❈❤ù♥❣ ♠✐♥❤ x5 + y + z ≥ x2 + y + z ❱➻ x ≥ −2 ♥➯♥ t❛ ❝â (x + 2)(x − P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ 1)2 ≥ ⇒ x3 ≥ 3x − 2✳ ❉♦ ✤â x5 ≥ 3x3 − 2x2 (1) ❚÷ì♥❣ tü t❛ ❝â y ≥ 3y − 2y (2) z ≥ 3z − 2z (3) ❈ë♥❣ ✈➳ t❤❡♦ ✈➳ ✭✶✮✱ ✭✷✮✱ ✭✸✮ t❛ ❝â x5 + y + z ≥ 3(x3 + y + z ) − 2(x2 + y + z ) ≥ x2 + y + z (✤✐➲✉ ✼ ♣❤↔✐ ❝❤ù♥❣ ♠✐♥❤) ... t❤✐➯♥✿ x −∞ = +∞✳ −2 − y x2 + − f +∞ + +∞ y −16 • • • ã 16 số ỗ tr (2; 0) ✈➔ (2; +∞)✳ ❍➔♠ sè ♥❣❤à❝❤ ❜✐➳♥ tr➯♥ ❝→❝ (; 2) (0; 2) ỗ t số ✤↕t ❝ü❝ ✤↕✐ t↕✐ (0; 0) ✈➔ ✤↕t ❝ü❝ t✐➸✉ t (2; 16) (2; 16)... (2y − 3x)i 13x2 − 32x − 21 = 2 x + y − = 2x + 3y ⇔ ⇔ 3x − y = 2y − 3x = −1 17 ⇔(x, y) ∈ (3; 4); − ; − 13 13 17 ❱➟② z = + 4i ❤♦➦❝ z = − 137 − 13 i✳ ❈➙✉ ✸✳ ●✐↔✐ ♣❤÷ì♥❣ tr➻♥❤ log3 (x + 1)2 +...✣⑩P ⑩◆ ✣➋ ❚❍■ ❚❍Û ❙➮ ✶✼ ❈➙✉ ✶✳ ❈❤♦ ❤➔♠ sè y = x4 − 8x2 ✭✶✮✳ ❛✮ ❑❤↔♦ st sỹ t ỗ t (C) số t ữỡ tr t t ợ ỗ t (C) t õ a y (a) = −4✳ P❤➙♥ t➼❝❤✲▲í✐ ❣✐↔✐✳ ❛✮ • • • • ❚➟♣ ①→❝ ✤à♥❤✿ R✳