Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! ! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! "! H4"I)1%J%)>K)L!ML)N=O9)P%E)Q)L4RS()/T&1)L4U&4)VE6) DW&()L"I&X)/Y)Z[)\.]^*) V1US)#4%)()*\]*7].*\^) L4_%)1%E&)$U6)`U%()\a*)@4b#c)24W&1)2d)#4_%)1%E&)1%E")>K) e%f&)4g)>0&1)23)24"I)489)Q)!"#$%&'()*+,-) ).*.)Q)B4%)#%C#()hhhF6E#4$%&2:FG&)) Bi=)\)j.c*)>%d6kF!#$%!$&'!()! y = 1 3 x 3 − 5m 2 x 2 − 4mx + 2 (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";!5<1! m = −1 *! =* >?'!'!7@!:";!AB!$C1!A/A!.D9! x 1 ,x 2 (C%!A$%! (x 1 2 + 5mx 2 +12m)(x 2 2 + 5mx 1 +12m) =1 *!! Bi=).)j\c*)>%d6kF) C; E1,1!0F.!G$HI3J!.D?3$! 1 log 2 (x 2 − x + 2) ≥ 1 log 2 (x +1) *! 0; >?'!J1-!.D9!K<3!3$F.!5&!3$L!3$F.!AMC!$&'!()! f (x) = (x +1).e x 2 −x !.D43!7%N3!OP"Q=R*!!! Bi=)7)j\c*)>%d6kF!>S3$!.SA$!G$T3! I = (x −1).cos 2 x 2 dx 0 π ∫ *! Bi=)l)j\c*)>%d6kF)) C; #$%!()!G$UA! z = 1+ i 3 *!V12.!W!XH<1!XN3J!KHY3J!J1-A*!>?'!3!3JZ[43!XHI3J!3$L!3$F.!7@! z n K&!()!3JZ[43!XHI3J*! 0; #$%!=3!71@'! (n ≥ 2,n ∈ !) K&!A-A!7\3$!AMC!'].!7C!J1-A!7^Z*!_12.!()!.C'!J1-A!5Z`3J!.N%! .$&3$!.a!=3!71@'!7B!0b3J!"cd*!>?'!3*!!! Bi=)^)j\c*)>%d6kF)#$%!$?3$!A$BG!e*f_#!AB!7-[!f_#!K&!.C'!J1-A!7^Z!AN3$!Cg! SA = a *!Eh1!igj! Kk3!KHY.!K&!.DZ3J!71@'!A-A!AN3$!_#!5&!ef*!E1,!(l!ei!5Z`3J!JBA!5<1!'m.!G$n3J!:f_#;*!>S3$! .$@!.SA$!o$)1!A$BG!e*f_#!5&!A`(13!JBA!J1pC!$C1!7Hq3J!.$n3J!_j!5&!f#*!!! Bi=)-)j\c*)>%d6kF)>D%3J!o$`3J!J1C3!5<1!$r!.DsA!.%N!7]!tu[W!A$%!$C1!71@'!f:"QdQv;g!_:wQ=Q=;!5&! 'm.!G$n3J! (P ) : x + y + z + 8 = 0 *!>S3$!o$%,3J!A-A$!.a!.DZ3J!71@'!AMC!7%N3!.$n3J!f_!723!'m.! G$n3J!:x;*!>?'!.%N!7]!71@'!i!.D43!:x;!7@! MA 2 + MB 2 3$L!3$F.*!!! Bi=),)j\c*)>%d6kF)>D%3J!'m.!G$n3J!5<1!.DsA!.%N!7]!tu[!A$%!$?3$!5Z`3J!f_#y!.T'!z*!Eh1!ig! jg{!Kk3!KHY.!K&!.DZ3J!71@'!A-A!7%N3!.$n3J!fzg!#yg_j*!E1,!(l!G$HI3J!.D?3$!7Hq3J!.$n3J!i{!K&! y − 7 2 = 0 5&!j:|Q};*!>?'!.%N!7]!A-A!7\3$!$?3$!5Z`3J!f_#y!012.!7\3$!#!AB!$%&3$!7]!K<3!$I3!~*!! Bi=)a)j\c*)>%d6kF!E1,1!$r!G$HI3J!.D?3$! x 2 −3y 2 + x + 4y − 2 = ( y −1) 2 +1 x y 2 −3x 2 − 2x − 2y + 2 = −2. x 2 + x y ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ (x, y ∈ !) *! Bi=)+)j\c*)>%d6kF!#$%!Cg0gA!K&!A-A!()!.$/A!XHI3J!.$%,!'•3! ab + bc + ca =1 *!>?'!J1-!.D9!K<3!3$F.! AMC!01@Z!.$UA! P = a 2 + bc a 2 + (b + c ) 2 + b 2 + ca b 2 + (c + a) 2 + c 2 + ab c 2 + (a +b) 2 − 8 3(a 2 + b 2 + c 2 + 2) 5 *! mmm!nLmmm) ) Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! ! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! =! PHÂN TÍCH BÌNH LUẬN VÀ ĐÁP ÁN Bi=)\)j.c*)>%d6kF!#$%!$&'!()! y = 1 3 x 3 − 5m 2 x 2 − 4mx + 2 (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";!5<1! m = −1 *! =* >?'!'!7@!:";!AB!$C1!A/A!.D9! x 1 ,x 2 (C%!A$%! (x 1 2 + 5mx 2 +12m)(x 2 2 + 5mx 1 +12m) =1 *!! "* €hA!(13$!./!J1,1*! =* >C!AB•! y ' = x 2 − 5mx − 4m; y ' = 0 ⇔ x 2 − 5mx − 4m = 0 *! ‚@!:";!AB!$C1!A/A!.D9!o$1!5&!A$\!o$1![ƒ!AB!$C1!3J$1r'!G$T3!01r.!! ! ⇔ Δ = 25m 2 +16m > 0 ⇔ m > 0 m <− 16 25 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ *! +$1!7B!!.$„%!51P….!AB! x 1 + x 2 = 5m 5&! x 1 2 − 5mx 1 − 4m = 0;x 2 2 − 5mx 2 − 4m = 0 *!! V?!5†[!! (x 1 2 + 5mx 2 +12m)(x 2 2 + 5mx 1 +12m) =1 ⇔ (x 1 2 − 5mx 1 − 4m +5m(x 1 + x 2 ) +16m)(x 2 2 − 5mx 2 − 4m +5m(x 1 + x 2 ) +16m) = 1 ⇔ (5m(x 1 + x 2 ) +16m) 2 = 1 ⇔ (25m 2 +16m) 2 = 1 ⇔ 25m 2 +16m =1(do25m 2 +16m > 0) ⇔ m = −8± 89 25 *!!!! oU%)#;@)#<p&1)#q)m)>?'!'!7@!:";!AB!$C1!A/A!.D9! x 1 ,x 2 (C%!A$%! A = m 2 x 1 2 + 5mx 2 +12m + x 2 2 + 5mx 1 +12m m 2 7N.!J1-!.D9!3$L!3$F.*!!!!! €y•!! A = m 2 (x 1 2 −5mx 1 − 4m)+ 5m(x 1 + x 2 ) +16m + (x 2 2 −5mx 2 − 4m)+ 5m(x 1 + x 2 ) +16m m 2 = m 5(x 1 + x 2 ) +16 + 5(x 1 + x 2 ) +16 m = m 5m +16 + 5m +16 m ≥ 2 m 5m +16 . 5m +16 m = 2 *! yFZ!0b3J!u,[!DC!o$1! m 5m +16 = 5m +16 m = 1 ⇔ 5m +16 = m ⇔ m = −4(t / m ) *! ‚‡(•! m = −4 *! Bi=).)j\c*)>%d6kF) C; E1,1!0F.!G$HI3J!.D?3$! 1 log 2 (x 2 − x + 2) ≥ 1 log 2 (x +1) *! 0; >?'!J1-!.D9!K<3!3$F.!5&!3$L!3$F.!AMC!$&'!()! f (x) = (x +1).e x 2 −x !.D43!7%N3!OP"Q=R*!!! C; ‚1^Z!o1r3•! x +1> 0 log 2 (x +1) ≠ 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ −1< x ≠ 0 *! >C!AB•! log 2 (x 2 − x + 2) = log 2 (x − 1 2 ) 2 + 7 4 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ≥ log 2 7 4 > 0 *!! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! ! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! ~! ˆ;!j2Z! −1< x < 0 ⇒ log 2 (x +1) < 0 *!_F.!G$HI3J!.D?3$!KZ`3!7‰3J*! ˆ;!j2Z! x > 0 ⇒ log 2 (x +1) > 0 *!!! _F.!G$HI3J!.D?3$!.HI3J!7HI3J!5<1•! ! log 2 (x 2 − x + 2) ≤ log 2 (x +1) ⇔ x 2 − x + 2 ≤ x +1 ⇔ x 2 − 2x +1≤ 0 ⇔ x =1 *! V†[!.†G!3J$1r'!AMC!0F.!G$HI3J!.D?3$!K&! S = (−1;0) ∪ 1 { } *!! oU%)#;@)#<p&1)#q)m!E1,1!0F.!G$HI3J!.D?3$! 1 log 2 (x 2 − 3x + 4) > 2 log 2 (x +1) *! ‚‡(•! S = (−1;0) ∪ (1;3) *!!!! 0; €&'!()!Š:u;!K143!.sA!.D43!7%N3!OP"Q=R*! >C!AB•! f '(x) = e x 2 −x + (x +1)(2x −1).e x 2 −x = (2x 2 + x).e x 2 −x *! ! f '(x) = 0 ⇔ 2x 2 + x = 0 ⇔ x = 0 x = − 1 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ *! ˆ;!>S3$!7HYA•! f (−1) = 0; f (− 1 2 ) = 1 2 .e 3 4 ; f (1) = 2; f (2) = 3e 2 *! V?!5†[! max x∈ −1;2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ f (x) = f (2) = 3e 2 ; min x∈ −1;2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ f (x) = f (−1) = 0 *!!!! Bi=)7)j\c*)>%d6kF!>S3$!.SA$!G$T3! I = (x −1).cos 2 x 2 dx 0 π ∫ *! >C!AB•!!! I = 1 2 (x −1)(1+ cos x )dx 0 π ∫ = 1 2 (x −1)dx 0 π ∫ K ! "##### $##### + 1 2 (x −1)cos x dx 0 π ∫ M ! "####### $####### *! ˆ;! K = 1 2 (x −1)dx 0 π ∫ = 1 2 ( x 2 2 − x ) π 0 = π 2 4 − π 2 *! ˆ;!! u = x −1 dv = cosxdx ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ du = dx v = sin x ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ M = (x −1)sin x π 0 − sin x dx 0 π ∫ = cosx π 0 = −2 *! HC#)$=;&()V†[! K = π 2 4 − π 2 −1 *! Bi=)l)j\c*)>%d6kF)) C; #$%!()!G$UA! z = 1+ i 3 *!V12.!W!XH<1!XN3J!KHY3J!J1-A*!>?'!3!3JZ[43!XHI3J!3$L!3$F.!7@! z n K&!()!3JZ[43!XHI3J*! 0; #$%!=3!71@'! (n ≥ 2,n ∈ !) K&!A-A!7\3$!AMC!'].!7C!J1-A!7^Z*!_12.!()!.C'!J1-A!5Z`3J!.N%! .$&3$!.a!=3!71@'!7B!0b3J!"cd*!>?'!3*!!! C; >C!AB•! z = 2( 1 2 + 3 2 .i ) = 2(cos π 3 + i.sin π 3 ) *! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! ! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! v! V?!5†[! z n = 2(cos π 3 + i.sin π 3 ) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ n = 2 n (cos nπ 3 + i.sin nπ 3 ) *! ‚@! z n K&!()!3JZ[43!XHI3J!o$1!5&!A$\!o$1! sin nπ 3 = 0 2 n .cos nπ 3 ∈ ! * ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⇔ nπ 3 = k2π ⇔ n = 6k,k ∈ " *! V?!3!3JZ[43!XHI3J!3$L!3$F.!343! k = 1 ⇒ n = 6 *! V†[! n = 6 K&!J1-!.D9!Ak3!.?'*!!!!!! 0; ˆ;!>C'!J1-A!5Z`3J!7HYA!.N%!.$&3$!.a!'].!7Hq3J!oS3$!AMC!7Hq3J!.D‹3!3]1!.12G!7C!J1-A!7^Z! 5&!"!7\3$!o$`3J!3b'!.D43!7Hq3J!oS3$*! ˆ;!#B!.F.!A,!3!7Hq3J!oS3$*!V<1!'Œ1!7Hq3J!oS3$!AB!:=3P=;!7\3$!A‹3!KN1!A•3J!5<1!7Hq3J!oS3$!7B! 7@!.N%!.$&3$!'].!.C'!J1-A!5Z`3J*! V†[!AB!.F.!A,! n.(2n − 2) = 2(n 2 − n) .C'!J1-A!5Z`3J*! >$„%!J1,!.$12.!.C!AB•! 2(n 2 −n) = 180 ⇔ n =10(t / m) n = −9(l ) ⎡ ⎣ ⎢ ⎢ *!! V†[! n =10 K&!J1-!.D9!Ak3!.?'*!! Bi=)^)j\c*)>%d6kF)#$%!$?3$!A$BG!e*f_#!AB!7-[!f_#!K&!.C'!J1-A!7^Z!AN3$!Cg! SA = a *!Eh1!igj! Kk3!KHY.!K&!.DZ3J!71@'!A-A!AN3$!_#!5&!ef*!E1,!(l!ei!5Z`3J!JBA!5<1!'m.!G$n3J!:f_#;*!>S3$! .$@!.SA$!o$)1!A$BG!e*f_#!5&!A`(13!JBA!J1pC!$C1!7Hq3J!.$n3J!_j!5&!f#*!!! ! ˆ;!>C'!J1-A!5Z`3J!efi!AB! SM = SA 2 − AM 2 = a 2 − 3a 2 4 = a 2 *! ˆ;! S ABC = 1 2 AM .BC = a 2 3 4 *! eZ[!DC•! V S .ABC = 1 3 SM .S ABC = 1 3 . a 2 . a 2 3 4 = a 3 3 24 !:75 ;*! ˆ;!>S3$!JBA!J1pC!_j!5&!f#•! Eh1!Ž!K&!.DZ3J!71@'!AN3$!e#*!>C!AB!jŽ‡‡f#!343!JBA!J1pC!f#!5&! _j!0b3J!JBA!J1pC!jŽ!5&!_j*! >C'!J1-A!_jŽ!AB•! ! NE = AC 2 = a 2 ;BN = 2(AB 2 + SB 2 )− SA 2 2 = a 2 2 Q BE = 2(SB 2 + BC 2 )− SC 2 2 = a 10 4 *! eZ[!DC! cos BNE ! = BN 2 + NE 2 − BE 2 2BN .NE = a 2 2 + a 2 4 − 5a 2 8 2. a 2 . a 2 2 = 2 8 *!V?!5†[! cos(BN ; AC ) ! = 2 8 *!!! BI94).(!‚m.!$r!.DsA!J8'!i:dQdQd;g! A( a 3 2 ;0;0),B (0;− a 2 ;0),C (0; a 2 ;0),S (0;0; a 2 ),N ( a 3 4 ;0; a 4 ) *! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! ! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! "! #$%!&'! cos(BN ; AC ) ! = BN " #"" .AC " #"" BN " #"" . AC " #"" = 2 8 (!!!!!! BH=)-)IJK*)>%L6MF))&*+,! /+,!,0'+!120!.3!4&56!4*7!89!:;%<!6.*!.'0!80=>!?@ABCBDEF!G@HBIBIE!1J! >K4!L.M+,! (P ) : x + y + z + 8 = 0 (!)N+.! *O+,!6P6.!4Q!4&$+,!80=>!6R'!8*7+!4.M+,!?G!8S+!>K4! L.M+,!@TE(!)U>!4*7!89!80=>!V!4&W+!@TE!8=! MA 2 + MB 2 +.X!+.Y4(!!! ZE![\0!]!^J!4&$+,!80=>!6R'!?G!4'!6_!]@DBAB`E(!)'!6_a! d (I ;(P )) = 4 +1+ 3+ 8 1+1+1 = 16 3 3 (!! ZE!).b*!6/+,!4.c6!8de+,!4&$+,!4$%S+!4'!6_! MA 2 + MB 2 = 2MI 2 + AB 2 2 (! fU!1g%! MA 2 + MB 2 +.X!+.Y4! 0!V]!+.X!+.Y4!F!80h$!+J%!4di+,!8di+,!120!V!^J!.U+.!6.0S$! 1$/+,!,_6!6R'!]!4&W+!@TE(! ZE!jde+,!4.M+,!k!80!l$'!]!1J!1$/+,!,_6!120!@TE!+.g+! n P !"! = (1;1;1) ^J>!1m6!4i!6.n!L.di+,!+W+!6_! L4!^J! d : x = 4 + t y = 1+ t z = 3+ t ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ (!).'%!;F%F<!4Q!L4!6R'!k!1J*!L.di+,!4&U+.!6R'!@TE!4'!8do6a! ! 4 + t +1+ t + 3+ t + 8 = 0 ⇔ t = − 16 3 ⇒ M − 4 3 ;− 13 3 ;− 7 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ (! fg%!80=>!6p+!4U>!^J! M − 4 3 ;− 13 3 ;− 7 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ (!! BH=),)IJK*)>%L6MF))&*+,!>K4!L.M+,!120!4&56!4*7!89!:;%!6.*!.U+.!1$/+,!?Gqr!4s>!](![\0!VF! tFu!^p+!^do4!^J!4&$+,!80=>!6P6!8*7+!4.M+,!?]F!qrFGt(![0O!vw!L.di+,!4&U+.!8de+,!4.M+,!Vu!^J! y − 7 2 = 0 1J!t@"BxE(!)U>!4*7!89!6P6!8n+.!.U+.!1$/+,!?Gqr!y0S4!8n+.!q!6_!.*J+.!89!^2+!.i+!`(!! ! )'!6.c+,!>0+.!Vu!1$/+,!,_6!120!tua! BN94)J(!rz+,!89!kJ0!67+.!.U+.!1$/+,F!8K4! AB = a > 0 (! rz+,!8{+.!^|!.J>!v}!6/v0+!6.*!6P6!4'>!,0P6!?GVB!qVtB!Gqt! )N+.!8do6a! MB = MN = a 10, BN = 2a 5 (!!! tW+!4'>!,0P6!GVt!1$/+,!6s+!470!V!v$%!&'! MJ ⊥ NJ (! qP6.!Ia!![\0!~!^J!4&$+,!80=>!8*7+!rt(! )'!6_a! JK //BD BD ⊥ AC ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ JK ⊥ MI (1); MK //AD IJ ⊥ AD ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ IJ ⊥ MK (2) (!! )Q!@AE!1J!@IE!v$%!&'!]!^J!4&•6!4s>!4'>!,0P6!Vu~(! fU!1g%! IK ⊥ MJ IK //NJ ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ NJ ⊥ MJ (!!! jde+,!4.M+,!tu!80!l$'!t!1J!1$/+,!,_6!120!Vu!6_!L.di+,!4&U+.!^J! x −5 = 0 (!)*7!89!80=>!u!^J! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! ! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! x! +,.03>!6R'!.3! x −5 = 0 y − 7 2 = 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇔ x = 5 y = 7 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇒ J (5; 7 2 ) (! fU!u!^J!4&$+,!80=>!Gt!+W+!G@"BAE(! [\0!q@'ByE!120!'€`!4'!6_! BC = 2NC = 2BN 5 = 2 5 (!!! )'!6_!.3!L.di+,!4&U+.a! (a − 5) 2 + (b −1) 2 = 20 (a − 5) 2 + (b −6) 2 = 5 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ (b −1) 2 −(b −6) 2 = 15 (a − 5) 2 + (b −1) 2 = 20 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ a = 7,b = 5(t / m) a = 3,b = 5(l) ⎡ ⎣ ⎢ ⎢ (! fU!1g%! C (7;5) F!k*!t!^J!4&$+,!80=>!qr!+W+!r@`BHE!1J! DC ! "!! = AB ! "!! ⇒ A(1;3) (! fg%!4*7!89!y}+!8n+.!6p+!4U>!^J A(1;3),B(5;1),C (7;5),D(3;7) (! BH=)O)IJK*)>%L6MF![0O0!.3!L.di+,!4&U+.! x 2 −3y 2 + x + 4y − 2 = ( y −1) 2 +1 x y 2 −3x 2 − 2x − 2y + 2 = −2. x 2 + x y ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ (! j0h$!-03+a! x ≠ 0; y ≠ 0 (! •3!L.di+,!4&U+.!4di+,!8di+,!120a! x 3 −3xy 2 + 4xy = y 2 − x 2 + 2(x − y +1) (1) y 3 −3x 2 y − 2xy = 2( y 2 − x 2 − x − y) (2) ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ (! ‚Y%!@AE!ƒ!@IE(0!!1J!8K4! z = x + y.i !4.b*!1S!4'!8do6a! ⇔ z 3 + (1− 2i)z 2 − 2z(1+ i) − 2 = 0 ⇔ (z +1)(z 2 − 2i.z −2) = 0 ⇔ (z +1)(z −1− i)(z +1−i) = 0 ⇔ z = −1 z = 1+ i z = −1+ i ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⇔ x + yi = −1 x + yi = 1+ i x + yi = −1+ i ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⇔ x = −1, y = 0 x =1, y =1 z = −1, y =1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ (! PC#)$=;&()fg%!.3!L.di+,!4&U+.!6_!.'0!+,.03>!^J! (x; y) = (1;1);(−1;1) (!!! B4Q)3F!)'!6_a! z 2 = x 2 − y 2 + 2xy.i;z 3 = x 3 − 3xy 2 + (−y 3 + 3x 2 y).i (!fU!1g%!kY$!.03+!vw!k5+,!v}! L.c6!^J!4&*+,!L.di+,!4&U+.!6_!6P6!870!^do+,! x 2 − y 2 ;x 3 − 3xy 2 ; y 3 − 3x 2 y (!! RS%)#;@)#<T&1)#U)V[0O0!.3!L.di+,!4&U+.! x 3 −3xy 2 − x −1 = y 2 + 2xy − x 2 y 3 −3x 2 y + y +1= x 2 + 2xy − y 2 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ (! BH=)+)IJK*)>%L6MF!q.*!'FyF6!^J!6P6!v}!4.•6!kdi+,!4.*O!>„+! ab + bc + ca =1 (!)U>!,0P!4&{!^2+!+.Y4! 6R'!y0=$!4.c6! P = a 2 + bc a 2 + (b + c ) 2 + b 2 + ca b 2 + (c + a) 2 + c 2 + ab c 2 + (a +b) 2 − 8 3(a 2 + b 2 + c 2 + 2) 5 (! Lời$giải:$ #w!k5+,!yY4!8M+,!4.c6!?V!…[V!4'!6_a! ! a 2 + bc a 2 + (b + c ) 2 ≤ a 2 + (b + c ) 2 4 a 2 + (b + c ) 2 = 1− 3 4 . (b + c ) 2 a 2 + (b + c ) 2 (! )di+,!4•!v$%!&'a! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! ! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! H! ! a 2 + bc a 2 + (b + c ) 2 ≤ 3− 3 4 . (b + c ) 2 a 2 + (b + c ) 2 ∑∑ (!! #w!k5+,!yY4!8M+,!4.c6!q'$6.%!…#6.†'&<!4'!6_a! ! (b + c ) 2 a 2 + (b + c ) 2 ≥ 4(a + b + c) 2 (a 2 + (b + c ) 2 ) ∑ = 4(a + b + c) 2 (a + b + c) 2 + 2(a 2 + b 2 + c 2 ) ∑ (! #$%!&'a! ! a 2 + bc a 2 + (b + c ) 2 ≤ 3− 3 4 . 4(a + b + c) 2 (a + b + c ) 2 + 2(a 2 + b 2 + c 2 ) ∑ = 6(a 2 + b 2 + c 2 ) (a + b + c ) 2 + 2(a 2 + b 2 + c 2 ) ≤ 6(a 2 + b 2 + c 2 ) (a + b + c ) 2 + 2 3 (a + b + c ) 2 = 18 5 . a 2 + b 2 + c 2 (a + b + c ) 2 (!! !! fU!1g%!! ! P ≤ 18 5 . a 2 + b 2 + c 2 (a + b + c) 2 − 8 3(a 2 + b 2 + c 2 + 2) 5 = 18 5 (1− 2 (a + b + c) 2 )− 8 3 5 (a + b + c) = f (t ) = 18 5 − 36 5t 2 − 8 3 5 t,t = a +b + c (! )'!6_a f '(t) = 72 5t 3 − 8 3 5 ; f '(t) = 0 ⇔ t = 3 (!fU!‡ˆ@4E!8‰0!kY$!kdi+,!v'+,!s>! 0!80!l$'! t = 3 (! #$%!&'! P ≤ f (t ) ≤ f ( 3) = 18 5 − 36 15 − 24 5 = − 18 5 (! fg%!,0P!4&{!^2+!+.Y4!6R'!T!yŠ+,! − 18 5 (!rY$!yŠ+,!8K4!470! a = b = c = 1 3 (!!!!! ! ! !! !! . Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! ! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! "! H4"I)1%J%)>K)L!ML)N=O9)P%E)Q)L4RS()/T&1)L4U&4)VE6) DW&()L"I&X)/Y)Z[).]^*) V1US)#4%)()*]*7].*^) L4_%)1%E&)$U6)`U%()a*)@4b#c)24W&1)2d)#4_%)1%E&)1%E")>K) e%f&)4g)>0&1)23)24"I)489)Q)!"#$%&'()*+,-). a) 2 + c 2 + ab c 2 + (a +b) 2 − 8 3(a 2 + b 2 + c 2 + 2) 5 *! mmm!nLmmm) ) Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! ! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! =! PHÂN. log 2 (x 2 − x + 2) = log 2 (x − 1 2 ) 2 + 7 4 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ≥ log 2 7 4 > 0 *!! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! ! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! ~! ˆ;!j2Z!