KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! "! Khoá%giải%đề%THPT%Quốc%Gia%–%Thầy:%Đặng%Thành%Nam% Môn:%Toán;%ĐỀ%SỐ%17/50% Ngày%thi%:%18/03/2015% Thời%gian%làm%bài:%180%phút,%không%kể%thời%gian%giao%đề% Liên%hệ%đăng%ký%khoá%học%–%Hotline:%0976%266%202%–%Chi%tiết:%www.mathlinks.vn%% Câu%1%(2,0%điểm).%#$%!$&'!()! y = x + 2− m x + m 2 (1) !*+,!'!-&!.$/'!()!.$012! m ≠1;m ≠ −2 3! "3 4$5%!(6.!(0!7,89!.$,:9!*&!*;!<=!.$>!$&'!()!?"@!*+,! m = −1 3! A3 BC'!'!<D!$&'!()!?"@!9E$>1$!7,89!.F:9!G$%59E! (−4;+∞) 3! Câu%2%(1,0%điểm).% /@ H,5,!I$JK9E!.FC9$! log 25 (x +1)+ 1 2 log 5 6x +1 =1 3! 7@ BC'!E,6!.F>!-+9!9$L.!*&!9$M!9$L.!1N/!$&'!()! y = log 2 (x +1)− x 3 !.F:9!<%O9!PQR"S3!! Câu%3%(1,0%điểm).!BT9$!.T1$!I$U9! I = x 3 −2x + (x −1)e x e x + x 2 dx 0 1 ∫ 3! Câu%4%(1,0%điểm).% /@ #$%!()!I$V1! z = cos x − i.sin x 3!BC'!()!.$01!W!(/%!1$%! z −i = 2 3! 7@ BC'!()!$O9E!1$V/! x 16 !.F%9E!G$/,!.F,D9! (x 2 − 1 x +1) n !7,8.!9!-&!()!.0!9$,:9!.$%5!'X9! C n 2 = n + 27 3! Câu%5%(1,0%điểm).!#$%!$C9$!1$YI!Z3[\#]!1Y!<6^![\#]!-&!$C9$!7C9$!$&9$!1Y! AB = a,AD = a 3, BAD ! = 30 0 3!_C9$!1$,8`!*`a9E!EY1!1N/!Z!-:9!'b.!I$c9E!?[\#]@!.Fd9E!*+,! .F`9E!<,D'!1O9$![]2!EY1!E,e/!Z#!*&!'b.!I$c9E!?[\#]@!7f9E! 60 0 3!BT9$!.$D!.T1$!G$),!1$YI! Z3[\#]!*&!G$%59E!161$!.g!<,D'![!<89!'b.!I$c9E!?Z\]@3! Câu%6%(1,0%điểm).!BF%9E!G$a9E!E,/9!*+,!$h!.Fi1!.%O!<j!kW^l!1$%!'b.!I$c9E! (P ) : 2x + y + 2z −14 = 0 3!B/'!E,61![\#!1U9!.O,![!1Y!\2#!.$`j1!'b.!I$c9E!?m@!*&!9$n9!H?oRpR"@! -&'!.Fq9E!.U'2!r?sRtRu"@!-&!.F`9E!<,D'!1O9$!\#3!BC'!.%O!<j!<,D'![3!v,8.!I$JK9E!.FC9$!<Jw9E! .$c9E!\#3! Câu%7%(1,0%điểm).!BF%9E!'b.!I$c9E!*+,!.Fi1!.%O!<j!kW^!1$%!./'!E,61!9$q9![\#! (AC > AB ) 3! Hq,! D(2;− 3 2 ) !-&!1$U9!<Jw9E!I$U9!E,61!.F%9E!EY1![2!x?u"RQ@!-&!'j.!<,D'!.$`j1!<%O9![#!.$%5! 'X9! AB = AE 3!BC'!.%O!<j!161!<y9$![2\2#!7,8.!I$JK9E!.FC9$!<Jw9E!.Fz9!9E%O,!.,8I!./'!E,61! [\#!-&! x 2 + y 2 + x − 2y −30 = 0 !*&![!1Y!$%&9$!<j!{JK9E3! Câu%8%(1,0%điểm).%H,5,!$h!I$JK9E!.FC9$ x + y 2 + x 2 + y 2 2 = 2 x 3 + y 3 2 3 8x 3 − 4 x + 3 + 6 = 3 x + y 2 ( x + y ) ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (x, y ∈ !) 3! Câu%9%(1,0%điểm).%#$%!/2721!-&!161!()!.$01!G$a9E!U'!.$%5!'X9! a 2 + b 2 + c 2 = 2 !*&! a ≥ b 3!BC'!E,6! .F>!-+9!9$L.!1N/!7,D`!.$V1! P = 1 a 3 + b 2 + c + 1 b 3 + c 2 + a − 5(a + c )(a + c −1) 4 KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! A! PHÂN%TÍCH%BÌNH%LUẬN%ĐÁP%ÁN% Câu%1%(2,0%điểm).%#$%!$&'!()! y = x + 2− m x + m 2 (1) !*+,!'!-&!.$/'!()!.$012! m ≠1;m ≠ −2 3! "3 4$5%!(6.!(0!7,89!.$,:9!*&!*;!<=!.$>!$&'!()!?"@!*+,! m = −1 3! A3 BC'!'!<D!$&'!()!?"@!9E$>1$!7,89!.F:9!G$%59E! (−4;+∞) 3! "3 _q1!(,9$!.0!E,5,3! A3 BnI!W61!<>9$|! D = !\ −m 2 { } 3!B/!1Y|! y ' = m 2 + m − 2 (x + m 2 ) 2 3! }!@!_&'!()!9E$>1$!7,89!.F:9!G$%59E! (−4;+∞) G$,!*&!1$y!G$,|! x + m 2 ≠ 0 y ' <0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ,∀x ∈ (−4;+∞) ⇔ m 2 + m − 2 < 0 −m 2 ∈ −4;+∞ ⎡ ⎣ ⎢ ) ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇔ −2 <m <1 3! Kết%luận:%vn^! −2 < m <1 -&!E,6!.F>!1~9!.C'3!! Câu%2%(1,0%điểm).% /@ H,5,!I$JK9E!.FC9$! log 25 (x +1)+ 1 2 log 5 6x +1 =1 3! 7@ BC'!E,6!.F>!-+9!9$L.!*&!9$M!9$L.!1N/!$&'!()! y = log 2 (x +1)− x 3 !.F:9!<%O9!PQR"S3!! /@ •,€`!G,h9|! x > − 1 6 3! m$JK9E!.FC9$!.JK9E!<JK9E!*+,|! 1 2 log 5 (x +1)+ 1 2 log 5 6x +1 =1 ⇔ log 5 (x +1)+ log 5 6x +1 = 2 ⇔ log 5 (x +1) 6x +1 = 2 ⇔ (x +1) 6x +1 = 25 ⇔ (x +1)( 6x +1 −5)+5(x −4) = 0 ⇔ 6(x +1)(x − 4) 6x +1 + 5 + 5(x −4) = 0 ⇔ (x − 4) 6(x +1) 6x +1 + 5 + 5 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = 0 ⇔ x = 4 3! Kết%luận:!m$JK9E!.FC9$!1Y!9E$,h'!{`^!9$L.! x = 4 3!! #$•!‚!1Y!.$D!.$01!$,h9!161!161$!G$61!(/`!<U^|! }@!_&'!()! y = log 25 (x +1)+ 1 2 log 5 6x +1 <=9E!7,89!9:9!I$JK9E!.FC9$!1Y!.),!</!'j.!9E$,h'2! 'b.!G$61!9$n9!.$L^!Wƒs!.$%5!'X9!I$JK9E!.FC9$3!vn^!9:9!I$JK9E!.FC9$!1Y!9E$,h'!{`^!9$L.! Wƒs3! }@!H,5,!I$JK9E!.FC9$|!! (x +1) 6x +1 = 25 ⇔ (x +1) 2 (6x +1) = 25 2 ⇔ (x − 4)(6x 2 + 37x +156) = 0 ⇔ x = 4 3! 7@ !_&'!()!<X!1$%!-,:9!.i1!.F:9!<%O9!PQR"S3! B/!1Y|! y ' = 1 (x +1).ln 2 −3x 2 ≤ 1 2ln 2 −3< 0 3! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! o! vn^!$&'!()!9E$>1$!7,89!.F:9!<%O9!PQR"S3! ]%!<Y! y min = y(1) = 0; y max = y(0) = 0 3!!! Câu%3%(1,0%điểm).!BT9$!.T1$!I$U9! I = x 3 −2x + (x −1)e x e x + x 2 dx 0 1 ∫ 3! I = x(x 2 + e x )−(2x + e x ) e x + x 2 dx 0 1 ∫ = (x − 2x + e x e x + x 2 )dx 0 1 ∫ = ( x 2 2 −ln e x + x 2 ) 1 0 = 1 2 −ln(e +1) 3! Kết%luận:!vn^! I = 1 2 − ln(e +1) 3!! Câu%4%(1,0%điểm).% /@ #$%!()!I$V1! z = cos x − i.sin x 3!BC'!()!.$01!W!(/%!1$%! z −i = 2 3! 7@ BC'!()!$O9E!1$V/! x 16 !.F%9E!G$/,!.F,D9! (x 2 − 1 x +1) n !7,8.!9!-&!()!.0!9$,:9!.$%5!'X9! C n 2 = n + 27 3! /@ B/!1Y|! z −i = cos x + (sin x +1).i ⇒ z − i = cos 2 x + (sin x +1) 2 3!! }@!B$„%!E,5!.$,8.!./!1Y|! ! cos 2 x + (sin x +1) 2 = 2 ⇔ cos 2 x + (sin x +1) 2 = 4 ⇔ 1−sin 2 x + (sin x +1) 2 = 4 ⇔ 2sin x + 2 = 4 ⇔ sin x = 1⇔ x = π 2 + k2π 3! 48.!-`n9|!vn^! x = π 2 + k2π,k ∈ ! 3!!! 7@ B$„%!E,5!.$,8.!./!1Y!I$JK9E!.FC9$|! ! n! 2!(n −2)! = n + 27 ⇔ n(n −1) 2 = n + 27 ⇔ n 2 −3n −54 = 0 ⇔ n = 9(t / m) n = −6(l ) ⎡ ⎣ ⎢ ⎢ 3! vn^! n = 9 *&! (x 2 − 1 x +1) 9 = C 9 k (x 2 − 1 x ) k k=0 9 ∑ = C 9 k . (−1) i x 2(k−i ) .x −i i=0 k ∑ k=0 9 ∑ = (−1) i .x 2k−3i i=0 k ∑ k=0 9 ∑ 3! B/!1~9!1$q9!G2,!.$%5!'X9! 0 ≤ i ≤ k ≤ 9 2k −3i = 16 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ (i;k ) = (0;8) 3! vn^!()!$O9E!1~9!.C'!-&! C 9 8 .(−1) 0 x 16 = 9x 16 3!! Câu%5%(1,0%điểm).!#$%!$C9$!1$YI!Z3[\#]!1Y!<6^![\#]!-&!$C9$!7C9$!$&9$!1Y! AB = a,AD = a 3, BAD ! = 30 0 3!_C9$!1$,8`!*`a9E!EY1!1N/!Z!-:9!'b.!I$c9E!?[\#]@!.Fd9E!*+,! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! s! .F`9E!<,D'!1O9$![]2!EY1!E,e/!Z#!*&!'b.!I$c9E!?[\#]@!7f9E! 60 0 3!BT9$!.$D!.T1$!G$),!1$YI! Z3[\#]!*&!G$%59E!161$!.g!<,D'![!<89!'b.!I$c9E!?Z\]@3! ! Hq,!_!-&!.F`9E!<,D'!1O9$![]2!./!1Y|! SH ⊥ (ABCD ) 3! v&!_#!-&!$C9$!1$,8`!*`a9E!EY1!1N/!Z#!.F:9!'b.!I$c9E! ?[\#]@!9:9! SCH ! = 60 0 3!! ]%! BAD ! = 30 0 ⇒ ADC ! = 150 0 3! …I!{i9E!<>9$!-‚!$&'!()!#a(,9!1$%!./'!E,61!_]#!1Y|! ! CH = HD 2 +CD 2 − 2HD.CD cos150 0 = a 2 4 + 3a 2 − 2. a 2 .a 3.(− 3 2 ) = a 19 2 3!! Z`^!F/|! SH = HC.tan 60 0 = a 19 2 . 3 = a 57 2 3!! }@! S ABCD = 2S ABD = AB.AD.sin30 0 = a.a 3. 1 2 = a 2 3 2 3! vC!*n^! V S .ABCD = 1 3 SH .S ABCD = 1 3 . a 57 2 . a 2 3 2 = a 3 19 4 ?<* @3! }@!BT9$!G$%59E!161$!.g![!<89!'b.!I$c9E!?Z\]@3! B/!1Y! d (A;(SBD)) = AD HD .d(H ;(SBD )) = 2.d(H ;(SBD)) 3! 4†!_‡!*`a9E!EY1!*+,!\]!.O,!‡!*&!G†!_4!*`a9E!EY1!*+,!Z‡!.O,!4!./!1Y! HK ⊥ (SBD ) 3! }@!…I!{i9E!<>9$!-‚!$&'!()!1a!(,9!1$%!./'!E,61![\]!1Y!! ! BD = AB 2 + AD 2 − 2AB.AD cos30 0 = a 2 + 3a 2 − 2a.a 3. 3 2 = a !3! B/!1Y! HI = 2S HBD BD = S ABD BD = S ABCD 2BD = a 2 3 4a = a 3 4 3! B/'!E,61!*`a9E!Z_‡!1Y|! ! 1 HK 2 = 1 SH 2 + 1 HI 2 = 16 3a 2 + 4 57a 2 = 308 57a 2 ⇒ HK = a 2 57 77 3! Z`^!F/|! d (A;(SBD)) = a 57 77 3!!!!!! Câu%6%(1,0%điểm).!BF%9E!G$a9E!E,/9!*+,!$h!.Fi1!.%O!<j!kW^l!1$%!'b.!I$c9E! (P ) : 2x + y + 2z −14 = 0 3!B/'!E,61![\#!1U9!.O,![!1Y!\2#!.$`j1!'b.!I$c9E!?m@!*&!9$n9!H?oRpR"@! -&'!.Fq9E!.U'2!r?sRtRu"@!-&!.F`9E!<,D'!1O9$!\#3!BC'!.%O!<j!<,D'![3!v,8.!I$JK9E!.FC9$!<Jw9E! .$c9E!\#3! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! "! #$!%&!'!()!*+, !*/0!*10!.234!567!-8-! AG ! "!! = 2GM ! "!! = (2;4;−4) ⇒ 3− x A = 2 6− y A = 4 1− z A = −4 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇒ A(1;2;5) 9! #$!:;!*10!.234!567!4/-!*<2!5!-8-! BC ⊥ AM 9! =1!4>?! AM ! "!!! = (3;6;−6) //(1;2;−2),n P !"! = (2;1;2) ⇒ AM ! "!!! ,n P !"! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = (6;−6;−3) //(2;−2;−1) 9! =1!4>! BC ⊥ AM ,BC ⊥ n P !"! !@AB!+1!67!-CD-!EFGHFGHI$!()0!JK4!*L!4CM!NCOL 9! :;!P>!67!P2!QA1!R!4>!NCOL !*+&-C!()! x − 4 2 = y −8 −2 = z +1 −1 9!!!! Kết%luận:!5EIGFG"$!J)! BC : x − 4 2 = y −8 −2 = z +1 −1 9! Câu%7%(1,0%điểm).!=+; !0S*!NCT !JU2!*+V4!*;<!PW!XYB!4C;!*10!.234!-C,-!567! (AC > AB ) 9! ',2! D(2;− 3 2 ) !()!4C/-!POZ !NC/-!.234!*+; !.>4!5[!\EHIG]$!()!0W*!P2^0!*CAW4!P;<-!57!*C;_! 0`-! AB = AE 9!=&0!*;<!PW!434!PM-C!5[6[7!a2b*!NCOL !*+&-C!POZ !*+c-! ;<2!*2bN!*10!.234! 567!()! x 2 + y 2 + x − 2y −30 = 0 !J)!5!4>!C;)-C!PW!dOL 9! ! eOZ !*+c-! ;<2!*2bN!*10!.234!567!4>!*/0! I (− 1 2 ;1) ![! a3-!fg-C!ah ! 5 5 2 9! ',2!i!()!.21;!P2^0!4j1!5k!J)!POZ !*CT !:\9! #$!lK*!C12!*10!.234!56:!J)!5\:!4>?! #! AB = AE,BAD ! = EAD ! [!5:!4CA !-8-!*10!.234!56:! ah !*10!.234!5:\9!! mAB!+1?! AED ! = ABC ! 9! =1!4>?! HAE ! = ICA ! = 180 0 − AIC ! 2 = 90 0 − ABC ! 9! mAB!+1?! AHE ! = AED ! + HAE ! = ABC ! + (90 0 − ABC ! ) = 90 0 9!%&!JDB!5k!JAn !.>4!JU2!POZ !*CT ! :\9!! #$!=1!4>?! DE ! "!! = (−3; 3 2 ) //(2;−1) 9!!!! #$!eOZ !*CT !5k!P2!QA1!k!J)!JAn !.>4!JU2!:\!-8-!4>!NCOL !*+&-C!()! 2x − y + 2 = 0 9! =;<!PW!P2^0!5!()! C2o0!4j1!Co! 2x − y + 2 = 0 x 2 + y 2 + x − 2y −30 = 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ x = 2, y = 6 x = −3,y = −4 ⎡ ⎣ ⎢ ⎢ ⇒ A(2;6) 9! #$!eOZ !*CT !5:!P2!QA1!5[:!4>!NCOL !*+&-C!()! x − 2 = 0 9! ',2!5p!()!.21;!P2^0!*Cq!C12!4j1!5:!J)!POZ !*+c-!E7$[!*;<!PW!P2^0!5p!*C;_!0`-! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! r! ! x = 2 x 2 + y 2 + x − 2y −30 = 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ x = 2, y = 6 x = 2, y = −4 ⎡ ⎣ ⎢ ⎢ ⇒ A'(2;−4) 9! #$!eOZ !*CT !67!P2!QA1!:!J)!JAn !.>4!JU2!k5p!4>!NCOL !*+&-C!()! ! x − 2 y −5 = 0 9! =;<!PW!P2^0!6[7!*C;_!0`-!Co! x −2y −5 = 0 x 2 + y 2 + x − 2y −30 = 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ x = 5, y = 0 x = −3, y = −4 ⎡ ⎣ ⎢ ⎢ 9! mAB!+1!6E"G]$[!7EHsGHt$!C;S4!6EHsGHt$[!7E"G]$9! eu2!4C2bA!JU2!P2vA!f2o-!57w56!*1!4>?!6E"G]$!J)!7EHsGHt$9! Kết%luận:!%DB! A(2;6),B(5;0),C (−3;−4) 9!! ! Câu%8%(1,0%điểm).%'2_2!Co!NCOL !*+&-C x + y 2 + x 2 + y 2 2 = 2 x 3 + y 3 2 3 8x 3 − 4 x + 3 + 6 = 3 x + y 2 ( x + y ) ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (x, y ∈ !) 9! e2vA!f2o-?! x, y ≥ 0 9! =1!4>?! ! x 2 + y 2 2 ≥ 1 4 (x + y) 2 = x + y 2 9! :;!P>! VT (1) ≤ 2 x 2 + y 2 2 ⇒VP (1) = 2 x 3 + y 3 2 3 ≤ 2 x 2 + y 2 2 9! mAB!+1?! ! ( x 3 + y 3 2 ) 2 ≤ ( x 2 + y 2 2 ) 3 ⇔ 2(x 3 + y 3 ) 2 −(x 2 + y 2 ) 3 ≤ 0 ⇔ (x − y) 2 (x 4 + 2x 3 y + 2xy 3 + y 4 ) ≤ 0 ⇔ x = y 9! :;!JDB!NCOL !*+&-C!PxA!4j1!Co!*OL !POL !JU2?! y = x 9! =C1B!J);!NCOL !*+&-C!*Cq!C12!4j1!Co!*1!POy4?! ! 4x 3 − 2 x + 3 + 3= 3x ⇔ 4x 3 − 3x −1+ 2(2− x + 3) = 0 ⇔ (x −1)(4x 2 + 4x +1)+ 2(1− x) 2+ x + 3 = 0 ⇔ (x −1)(4x 2 + 4x +1− 2 2+ x + 3 ) = 0 ⇔ (x −1)(8x 2 + 8x + (2x +1) 2 x + 3) = 0 ⇔ x = 1 9! 6z2!J&! 8x 2 + 8x + (2x +1) 2 x + 3 > 0,∀x ≥ 0 9! Kết%luận:!io!NCOL !*+&-C!4>! C2o0!dAB!-C{*! (x; y) = (1;1) 9!!!!!! Cách%2:!|CD-!*C{B!NCOL !*+&-C!PxA!4j1!Co!4>!d< !PT !4{N!-8-!*1!4>!*C^!PS*!B}*9Y!PO1!Jv! .2_2!NCOL !*+&-C!JU2!*9! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! ~! |CD-!*C{B! x = 0 fCn !()! C2o0!4j1!Co!-8-!YK*!JU2! x > 0 PS*! t = y x ≥ 0 NCOL !*+&-C!*Cq!-C{*! 4j1!Co!*+z!*C)-C?!! =1!4>?! ! (t 3 +1)(t 3 +1)(1+1) ≥ (t 2 +1) 3 ⇒ t 3 +1 2 3 ≥ t 2 +1 2 ≥ t +1 2 ⇒VT ≤VP 9! :{A!ah !Y_B!+1!fC2!J)!4CM!fC2! t = 1 9!! %&!JDB!NCOL !*+&-C!PxA!4j1!Co!*1!4>?! y = x 9! Bình%luận:!=1!4>!*C^!@AB!+1!a{*!PT !*Cq4!*+8-!-CO!@1A?! ! 2(t 3 +1) 2 −(t 2 +1) 3 = (t −1) 2 (t 4 + 2t 3 + 2t +1) ≥ 0,∀t ≥ 0 9! :;!P>! t 3 +1 2 3 ≥ t 2 +1 2 9!!!! #$!=• !QA3*!*1!4>?! ! (t n +1)(t n +1)(1+1) (1+1) (n−2) lân ! "##### $##### ≥ (t 2 +1) n ⇒ t n +1 2 n ≥ t 2 +1 2 9! :;!P>!NCOL !*+&-C!PxA!4j1!Co!4>!*C^!4C;!d< !*• !QA3*?! ! x + y 2 + x 2 + y 2 2 = 2. x n + y n 2 n 9! Bài%tập%tương%tự%}%% Bài%số%01.%'2_2!Co!NCOL !*+&-C! x + y 2 + x 2 + y 2 2 = 2 x 4 + y 4 2 4 x 3 −3x −1 = 8 − 3y 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ 9!! Câu%9%(1,0%điểm).%7C;!1[a[4!()!434!@u!*C€4!fCn !/0!*C;_!0`-! a 2 + b 2 + c 2 = 2 !J)! a ≥ b 9!=&0!.23! *+•!(U-!-C{*!4j1!a2^A!*Cq4! P = 1 a 3 + b 2 + c + 1 b 3 + c 2 + a − 5(a + c )(a + c −1) 4 9! ! m‚!dV !a{*!PT !*Cq4!71A4CB!ƒm4C„1+…!*1!4>?! (a 3 + b 2 + c )(a + b 2 + c 3 ) ≥ (a 2 + b 2 + c 2 ) 2 , ⇒ 1 a 3 + b 2 + c ≤ a + b 2 + c 3 (a 2 + b 2 + c 2 ) 2 ; (b 3 + c 2 + a)(b + c 2 + a 3 ) ≥ (a 2 + b 2 + c 2 ) 2 , ⇒ 1 b + c 2 + a 3 ≤ b + c 2 + a 3 (a 2 + b 2 + c 2 ) 2 9! %&!JDB! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! †! ! 1 a 3 + b 2 + c + 1 b 3 + c 2 + a ≤ a + b 2 + c 3 (a 2 + b 2 + c 2 ) 2 + b + c 2 + a 3 (a 2 + b 2 + c 2 ) 2 = a + b + b 2 + c 2 + a 3 + c 3 4 ≤ a + b + (b + c ) 2 + (a + c ) 3 4 ≤ a + b + (a + c) 2 + (a + c ) 3 4 9! mAB!+1?! ! P ≤ a + b + (a + c) 2 + (a + c ) 3 4 − 5(a + c )(a + c −1) 4 = (a + c) 3 − 4(a + c ) 2 + 5(a + c ) + a +b 4 9! =1!4>?! a + b ≤ 2(a 2 + b 2 ) ≤ 2(a 2 + b 2 + c 2 ) = 2 9! %)!PS*! x = a + c ⇒ f (x ) = x 3 − 4x 2 + 5x ≤ 2 9! %&!JDB! P ≤ a + b + f (x) 4 ≤ 2+ 2 4 = 1 9!:{A!ah !P<*!*<2! a = b = 1,c = 0 9! %DB!.23!*+•!(U-!-C{*!4j1!‡!ah !I9!!!!! 6h !434C!*OL !*€!*1!4Cq !02-C!POy4?! Bài%số%01.%7C;!1[a[4!()!434!@u!*C€4!dOL !*C;_!0`-! a + b + c = 3 9!7Cq !02-C!+h ! ! 1 a 3 + b 2 + c + 1 b 3 + c 2 + a + 1 c 3 + a 2 + b ≤1 9! ! !!!! ! ! !! . KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%– %Thầy: %ĐẶNG%THÀNH %NAM% % Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! "! Khoá %giải %đề% THPT%Quốc%Gia%– %Thầy: %Đặng %Thành% Nam% Môn: %Toán; %ĐỀ%SỐ%17/50% Ngày %thi% :%18/03/2015% Thời%gian%làm%bài:%180%phút,%không%kể%thời%gian%giao %đề% Liên%hệ%đăng%ký%khoá%học%–%Hotline:%0976%266%202%–%Chi%tiết:%www.mathlinks.vn%% Câu%1%(2,0%điểm).%#$%!$&'!()! . (1;1) 9!!!!!! Cách% 2:!|CD-!*C{B!NCOL !*+&-C!PxA!4j1!Co!4>!d< !PT !4{N!-8-!*1!4>!*C^!PS*!B}*9Y!PO1!Jv! .2_2!NCOL !*+&-C!JU2!*9! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%– %Thầy: %ĐẶNG%THÀNH %NAM% % Hotline:%0976%266%202%%. b 3!BC'!E,6! .F>!-+9!9$L.!1N/!7,D`!.$V1! P = 1 a 3 + b 2 + c + 1 b 3 + c 2 + a − 5(a + c )(a + c −1) 4 KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%– %Thầy: %ĐẶNG%THÀNH %NAM% % Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! A! PHÂN%TÍCH%BÌNH%LUẬN%ĐÁP%ÁN% Câu%1%(2,0%điểm).%#$%!$&'!()!