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 = x 3 − x 2 2 − 2x − 2 (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*! <* =>'!?!7@!7AB3C!.$D3C!E!FG!$H!()!CGF!?!5&!71!IJK!71@'!F/F!.1@J!FLK!:";M!FN.!:";!.O1!0K!71@'! P$Q3!01H.*! Bi=).)j\c*)>%d6kF) K; R1,1!P$AS3C!.T>3$! sin x.(cot 2 x − cot x ) = 3(cos x − sin x ) *! 0; !R1,1!P$AS3C!.T>3$! 2 x 2 +x−1 − 2 x 2 −1 = 2 2 x − 2 x *!! Bi=)7)j\c*)>%d6kF!=U3$!.UF$!P$Q3! I = cos2x 1+ 3(sin x −cos x ) 2 dx 0 π 2 ∫ *! Bi=)l)j\c*)>%d6kF) K; #$%!()!P$VF!W!.$%,!'X3! z − i.z 1+ i = −1+ 5i 2 *!=>'!()!P$VF!Y143!$ZP!FLK!()!P$VF! w = (1− 2i).z 2 *!! 0; R[1!\!Y&!.]P!$ZP!F-F!()!./!3$143!C8'!^!F$_!()!?$-F!3$KJ*!#$[3!3C`J!3$143!'a.!()!.b!\M! .U3$!c-F!(Jd.!7@!F$[3!7AZF!'a.!Ye3!$S3!<f"g*! Bi=)])j\c*)>%d6kF!#$%!$>3$!F$GP!h*\i#j!FG!7-k!\i#j!Y&!$>3$!F$_!3$].! AB = a,AD = 2a *! R[1!lMm!Yn3!YAZ.!Y&!.TJ3C!71@'!F-F!FO3$!\jMh#*!i12.!hl!5Jo3C!CGF!5e1!'p.!P$D3C!:\i#j;*! RGF!C1_K!'p.!P$D3C!:hij;!5&!'p.!P$D3C!:\i#j;!0q3C! 60 0 *!=U3$!.$@!.UF$!?$)1!F$GP!h*\i#j! 5&!?$%,3C!F-F$!C1_K!$K1!7AB3C!.$D3C!lm!5&!hi*!! Bi=)-)j\c*)>%d6kF!=T%3C!?$o3C!C1K3!5e1!$H!.TrF!.%O!7a!sckW!F$%!$K1!7AB3C!.$D3C! d : x −1 m 2 = y − 2 −n = z 4 ;Δ : x −m 1 = y −2 = z −1 1 !5e1! m,n ≠ 0 *!=>'!'M3!7@!$K1!7AB3C!.$D3C! d,Δ (%3C!(%3C!5e1!3$KJ!?$1!7G!512.!P$AS3C!.T>3$!'p.!P$D3C!:t;!F$VK! d,Δ *!!!! Bi=),)j\c*)>%d6kF!=T%3C!'p.!P$D3C!5e1!.TrF!.%O!7a!sck!F$%!$>3$!0>3$!$&3$!\i#j!FG!\:uvw";! 5&! AB ⊥ BD *!R[1!m!Y&!71@'!7)1!cV3C!FLK!#!IJK!j!5&! H ( 13 5 ; 9 5 ) Y&!$>3$!F$12J!5Jo3C!CGF!FLK!m! .T43!i#*!=>'!.%O!7a!F-F!7x3$!iM#Mj!012.!j!.$JaF! x + 2y −1 = 0 *!!! Bi=)a)j\c*)>%d6kF!R1,1!$H!P$AS3C!.T>3$! y x + 2 − x y + 2 = 2(x 3 − y 3 ) x 2(x + 1 y ) =1+ 2− y x 3 3 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ (x, y ∈ !) *! Bi=)+)j\c*)>%d6kF!#$%!cMkMW!Y&!F-F!()!.$/F!?$o3C!Q'!.$%,!'X3! x 2 + y 2 + z 2 = 2 *!=>'!C1-!.T9!3$y! 3$d.!FLK!01@J!.$VF! P = 1 x 2 + y 2 + 1 y 2 + z 2 + 1 z 2 + x 2 + 12 2− (x + y + z) 2 *! mmm!nLmmm) Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! <! M!oV)LpB!)qrV!)estV)/uM)uV) ! ! Bi=)\)j.c*)>%d6kF)#$%!$&'!()! y = x 3 − x 2 2 − 2x − 2 (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*! <* =>'!?!7@!7AB3C!.$D3C!E!FG!$H!()!CGF!?!5&!71!IJK!71@'!F/F!.1@J!FLK!:";M!FN.!:";!.O1!0K!71@'! P$Q3!01H.*! "* z[F!(13$!./!C1,1*! <* =K!FG!71@'!F/F!.1@J!FLK!:";!Y& A(1;− 7 2 ) *! {AB3C!.$D3C!E!71!IJK!\!$H!()!CGF!?!FG!EO3C|! y = k(x −1)− 7 2 *! t$AS3C!.T>3$!$%&3$!7a!C1K%!71@'|! ! x 3 − x 2 2 − 2x − 2 = k(x −1)− 7 2 ⇔ (x −1)(2x 2 + x − 3−2k) = 0 ⇔ x =1 2x 2 + x − 3−3k = 0 (*) ⎡ ⎣ ⎢ ⎢ *! {@!E!FN.!:";!.O1!0K!71@'!P$Q3!01H.!?$1!:};!FG!$K1!3C$1H'!P$Q3!01H.!?$-F!"*! ! ⇔ −3k ≠ 0 Δ = 1− 8(−3− 3k) > 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ − 25 24 < k ≠ 0 !*! HC#)$=;&(!~]k! − 25 24 < k ≠ 0 *!! Bi=).)j\c*)>%d6kF) F; R1,1!P$AS3C!.T>3$! sin x (cot 2 x − cot x ) = 3(cos x − sin x ) *! E; !R1,1!P$AS3C!.T>3$! 2 x 2 +x−1 − 2 x 2 −1 = 2 2 x − 2 x *!! K; {1•J!?1H3|! sin x ≠ 0 ⇔ x ≠ kπ,k ∈ ! *! t$AS3C!.T>3$!.AS3C!7AS3C!5e1|! ! cot 2 x −cot x = 3(cot x −1) ⇔ cot 2 x −( 3 +1) cot x + 3 = 0 ⇔ cot x =1 cot x = 3 ⎡ ⎣ ⎢ ⎢ ⎢ ⇔ x = π 4 + kπ x = π 6 + kπ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ *! ~]k!P$AS3C!.T>3$!FG!3C$1H'!Y&! x = π 4 + kπ, x = π 6 + kπ,k ∈ ! *!!! 0; t$AS3C!.T>3$!.AS3C!7AS3C!5e1|! ! (2 x 2 −1 − 2 x )(2 x −1) = 0 ⇔ 2 x 2 −1 = 2 x 2 x = 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⇔ x = 0 x 2 −1 = x ⎡ ⎣ ⎢ ⎢ ⇔ x = 0 x = 1± 5 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ *! ~]k!P$AS3C!.T>3$!FG!3C$1H'!Y&! x = 0;x = 1± 5 2 *!! qU%)#;@)#<v&1)#w)m!R1,1!P$AS3C!.T>3$! 2 2 x 2 −x − 2 x 2 −16.2 x 2 −x +16 = 0 *! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! v! {€(| x = 0;x = 1; x = −2;x = 2 *!!!! Bi=)7)j\c*)>%d6kF!=U3$!.UF$!P$Q3! I = cos2x 1+ 3(sin x −cos x ) 2 dx 0 π 2 ∫ *! =K!FG|! I = cos2x 4 −3sin 2x dx 0 π 2 ∫ *! {p.! t = 4 − 3sin 2x ⇒ t 2 = 4 − 3sin 2x ⇒ 2tdt = −6 cos 2xdx *! ~>!5]k I = − 1 3 tdt t −1 1 ∫ = − 1 3 t 1 −1 = − 2 3 *! Bi=)l)j\c*)>%d6kF) K; #$%!()!P$VF!W!.$%,!'X3! z − i.z 1+ i = −1+ 5i 2 *!=>'!()!P$VF!Y143!$ZP!FLK!()!P$VF! w = (1− 2i).z 2 *!! 0; R[1!\!Y&!.]P!$ZP!F-F!()!./!3$143!C8'!^!F$_!()!?$-F!3$KJ*!#$[3!3C`J!3$143!'a.!()!.b!\M! .U3$!c-F!(Jd.!7@!7AZF!'a.!Ye3!$S3!<f"g*! K; R1,!(•! z = x + yi(x, y ∈ !) .K!FG|! ! x + yi − i(x − yi) 1+ i = −1+ 5i 2 ⇔ x + yi − y + xi 1+ i = −1+ 5i 2 ⇔ x + yi − ( y + xi)(1− i ) 2 = −1+ 5i 2 ⇔ x + yi − x + y + (x − y)i 2 = −1+ 5i 2 ⇔ x − y + (3y − x )i 2 = −1+ 5i 2 ⇔ x − y = −1 3y − x = 5 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ x =1 y = 2 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ z = 1+ 2i *! ~>!5]k! w = (1+ 2i ) 2 (1− 2i ) = (1+ 2i ). (1+ 2i )(1−2i ) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 3(1+ 2i ) = 3+ 6i ⇒ w = 3−6i *! HC#)$=;&(! w = 3− 6i *!!!! 0; la.!()!.$JaF!\!FG!EO3C|! abcd *! ‚;! a ∈ 1,2,3,4,5,6,7,8,9 { } ⇒ a FG!ƒ!F-F$!F$[3*! ‚;! bcd FG! A 9 3 !F-F$!F$[3*! ~]k!.T%3C!\!FG!.d.!F,! 9.A 9 3 = 4536 ()*! R[1!„!Y&!0123!F)!F$[3!7AZF!'a.!()!.b!\!Ye3!$S3!<f"g*! ‚;!+$o3C!C1K3!'`J|! Ω = 4536 *! ‚;!=K!.>'!()!?I!.$J]3!YZ1!F$%!„!0q3C!F-F!.>'!F-F!()!?$o3C!5AZ.!IJ-!<f"g*! abcd ≤ 2015 ⇒ a ∈ 1,2 { } *! =z"|!m2J! a =1 M!FG!'a.!F-F$!F$[3!KM! bcd FG! A 9 3 F-F$!F$[3*! ~]k!.TAB3C!$ZP!3&k!FG! A 9 3 ()*! =z<|!m2J! a = 2 ⇒ abcd ∈ 2013,2014,2015 { } FG!v!()*! ~]k! Ω X = Ω −(3+ A 9 3 ) = 4029 *! 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&! ^! HC#)$=;&(!„-F!(Jd.!Fn3!.U3$!Y& P = 4029 4536 *!!!!!!!!! Bi=)])j\c*)>%d6kF!#$%!$>3$!F$GP!h*\i#j!FG!7-k!\i#j!Y&!$>3$!F$_!3$].! AB = a,AD = 2a *! R[1!lMm!Yn3!YAZ.!Y&!.TJ3C!71@'!F-F!FO3$!\jMh#*!i12.!hl!5Jo3C!CGF!5e1!'p.!P$D3C!:\i#j;*! RGF!C1_K!'p.!P$D3C!:hij;!5&!'p.!P$D3C!:\i#j;!0q3C! 60 0 *!=U3$!.$@!.UF$!?$)1!F$GP!h*\i#j! 5&!?$%,3C!F-F$!C1_K!$K1!7AB3C!.$D3C!lm!5&!hi*!! ! R[1!zM+!Yn3!YAZ.!Y&!$>3$!F$12J!5Jo3C!CGF!FLK!\Ml!.T43! ij*! =K!FG|! MK AH = DM DA = 1 2 *! =K'!C1-F!5Jo3C!\ij!FG!! 1 AH 2 = 1 AB 2 + 1 AD 2 = 1 a 2 + 1 4a 2 ⇒ AH = 2a 5 5 ;MK = a 5 5 *!! ‚;!j%!ij!5Jo3C!CGF!5e1!'p.!P$D3C!:hl+;!343! SKM ! = 60 0 *! hJk!TK|! SM = MK .tan 60 0 = a 5 5 . 3 = a 15 5 *! ~>!5]k! V S .ABCD = 1 3 SM .AB.AD = 1 3 . a 15 5 .a.2a = 2a 3 15 15 :75 ;*! ‚;!=U3$!?$%,3C!F-F$!C1_K!lm!5&!hi*! …]P!$H!.TrF!.%O!7a!FG!C)F! M (0;0;0),A(0;a;0),D(0;−a;0),B (a;a;0),C (a;−a;0),S (0;0; a 15 5 ) *! =K!FG!m!Y&!.TJ3C!71@'!FLK!h#!343!! N ( a 2 ;− a 2 ; a 15 10 ) *! =K!FG|! MN ! "!!! = ( a 2 ;− a 2 ; a 15 10 ),SB ! "! = (a;a;− a 15 5 ),MB ! "!! = (a;a;0) *! ~>!5]k! MN ! "!!! ,SB ! "! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = (0; a 2 15 5 ,a 2 );d (MN ,SB ) = MN ! "!!! ,SB ! "! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ .MB ! "!! MN ! "!!! ,SB ! "! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = a 3 15 5 a 4 + 15a 4 25 = a 6 4 *! HC#)$=;&(!!! V S .ABCD = 2a 3 15 15 ;d (MN ,SB ) = a 6 4 *!!!! Bi=)-)j\c*)>%d6kF!=T%3C!?$o3C!C1K3!5e1!$H!.TrF!.%O!7a!sckW!F$%!$K1!7AB3C!.$D3C! d : x −1 m 2 = y − 2 −n = z 4 ;Δ : x −m 1 = y −2 = z −1 1 !5e1! m,n ≠ 0 *!=>'!'M3!7@!$K1!7AB3C!.$D3C! d,Δ (%3C!(%3C!5e1!3$KJ!5&!?$1!7G!512.!P$AS3C!.T>3$!'p.!P$D3C!:t;!F$VK! d,Δ *!!!! {AB3C!.$D3C!E!71!IJK!71@'!l:"w<wf;!3$]3! a ! = (m 2 ;−n;4) Y&'!5†F!.S!F$x!P$AS3C*! {AB3C!.$D3C! Δ !!71!IJK!71@'!m:'wfw";!3$]3! b ! = (1;−2;1) Y&'!5†F!.S!F$x!P$AS3C*! j%!E€€ Δ !343! a ! ,b ! !F‡3C!P$AS3C!5&! MN ! "!!! ,b " !?$o3C!F‡3C!P$AS3C*! 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&! "! #$!!! a ! ,b ! !%&'(!)*+,'( ⇔ m 2 1 = −n −2 = 4 1 ⇔ m = ±2 n = 8 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ -! #$!! MN ! "!!! ,b " !.*/'(!%&'(!)*+,'(! ⇔ 1− m 1 ≠ 2 −2 ≠ −1 1 ⇔ m ≠ 2 -! 012!*3)!*45!6578!.59'!2:;'!24!%<!=>?@A'>B-! 0*5!6<! MN ! "!!! = (−3;−2;1) ⇒ MN ! "!!! ,b " ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = (0;4;8) / /(0;1;2) -! CD2!)*E'(!FG$!%*H4!IA! Δ !'*J'!FKLML@$!NO=!PQ%!2,!)*R)!28S1'!PO!65!T84!C!';'!%<!)*+,'(! 2:U'*!NO! (P ) : y + 2z − 2 = 0 -! HC#)$=;&(!=>?@A'>B!PO!FG$V!S#@W!?@>K-! BI=),)JKL*)>%M6NF!X:Y'(!=D2!)*E'(!PZ5!2:[%!2Y\!6]!^_S!%*Y!*U'*!`U'*!*O'*!abcd!%<!aF?eLM$! PO! AB ⊥ BD -!fg5!h!NO!65i=!6j5!_H'(!%k4!c!T84!d!PO! H ( 13 5 ; 9 5 ) NO!*U'*!%*518!P8/'(!(<%!%k4!h! 2:;'!bc-!XU=!2Y\!6]!%R%!6l'*!bAcAd!`512!d!2*8]%!6+m'(!2*E'(! x + 2y −1 = 0 -!!! ) ON)P4Q#)4%R&)#A&4)94S#)G=T&1)159(!X4!%*H'(!=5'*!an! P8/'(!(<%!PZ5!dn-! dY!ab!P8/'(!(<%!PZ5!bd!';'!2H!(5R%!abdh!NO!*U'*! %*o!'*J2!FM$-! p8S!:4! NAB ! = NHB ! = 90 0 PU!PJS!2H!(5R%!abnh!']5! 251)-!q+m'(!2:r'!'OS!%*s'*!NO!6+m'(!2:r'!'(Y\5!251)! 24=!(5R%!abh!%<!2t=!NO!2:8'(!65i=!%k4!bh!F@$-! Xu!FM$!PO!F@$!v8S!:4V!"!65i=!aAbAnAdAh!%&'(!2*8]%! 6+m'(!2:r'!6+m'(!.s'*!bh-! wU!PJS! AHD ! = AND ! = 90 0 A!2H%!an!P8/'(!(<%!PZ5!dn! F6)%=$-! ) ON)UV6)#"W)>X)>%M6())) G*+,'(!2:U'*!6+m'(!2*E'(!dn!65!T84!n!PO!P8/'(!(<%!PZ5!an!%<!)*+,'(!2:U'*! NO 7x + y − 20 = 0 -!! XY\!6]!65i=!d!NO!'(*59=!%k4!*9! 7x + y − 20 = 0 x + 2y −1= 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ x = 3 y = −1 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ D(3;−1) -! #$!q+m'(!2*E'(!bc!65!T84!n!PO!vY'(!vY'(!PZ5!ad!';'!%<!)*+,'(!2:U'*!NO! x + 3y − 8 = 0 -! fg5!bFB?e`L`$!PZ5!`!.*R%!xy"!2*8]%!6+m'(!2*E'(!bc!24!%<! AB ! "!! = (11− 3b;b −1),DB ! "!! = (5− 3b;b +1) -! X4!%<V! ! AB ! "!! .DB ! "!! = 0 ⇔ (11− 3b)(5−3b)+ (b −1)(b +1) = 0 ⇔ b = 3(t / m);b = 9 5 (l ) -! dY!6<!bF?MLe$-! wU! AB ! "!! = DC ! "!! ⇒ C (5;1) -! HC#)$=;&(!wJS!bF?MLe$A!cF"LM$!PO!dFeL?M$-!! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! z! BI=)Y)JKL*)>%M6NF!f5{5!*9!)*+,'(!2:U'*! y x + 2 − x y + 2 = 2(x 3 − y 3 ) x 2(x + 1 y ) =1+ 2− y x 3 3 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ (x, y ∈ !) -! q578!.59'V! x, y ≥ −2;x, y ≠ 0;x + 1 y ≥ 0 -! #$!h*J'!2*|S! x = −2 *YD%! y = −2 .*/'(!2*Y{!=}'!*9!)*+,'(!2:U'*-! #$!~Q2!PZ5! x > −2; y >−2 )*+,'(!2:U'*!2*H!'*|2!%k4!*9!2+,'(!6+,'(!PZ5V! y x + 2 − x y + 2 x + 2. y + 2 = 2(x 3 − y 3 ) x + 2. y + 2 ⇔ y y + 2 − x x + 2 = 2(x 3 − y 3 ) x + 2. y + 2 (1) -! ~Q2!*O=!vj! f (t) = t t + 2 ; f '(t) = t + 2 − t 2 t + 2 t + 2 = t + 4 2 (t + 2) 3 > 0 -! wU!PJS!'18! x > y ⇒VT 1 (x ) = f ( y) − f (x ) < 0;VP (1) > 0 )*+,'(!2:U'*!P/!'(*59=-! h18! x < y ⇒VT (1) = f (y)− f (x ) > 0;VP (1) < 0 )*+,'(!2:U'*!P/!'(*59=-! h*J'!2*|S! x = y 2*Y{!=}'-!X*4S!POY!)*+,'(!2:U'*!2*H!*45!%k4!*9!24!6+3%V!! ⇔ 2x 3 + 2x − 2 = 2 x 3 − 1 x 2 3 −1 ⇔ 2x 3 + 2x − 4 2x 3 + 2x + 2 = −x 3 − x + 2 x 3 ( 2 x 3 − 1 x 2 3 ) 2 + 1 x . 2 x 3 − 1 x 2 3 + 1 x 2 ⇔ (x 3 + x − 2) 2 2x 3 + 2x + 2 + 1 x 3 . 1 ( 2 x 3 − 1 x 2 3 ) 2 + 1 x . 2 x 3 − 1 x 2 3 + 1 x 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = 0 ⇔ x 3 + x − 2 = 0 ⇔ (x −1)(x 2 + x + 2) = 0 ⇔ x = 1 -! b•5!PU! 2 2x 3 + 2x + 2 + 1 x 3 . 1 ( 2 x 3 − 1 x 2 3 ) 2 + 1 x . 2 x 3 − 1 x 2 3 + 1 x 2 > 0(do x>0) -!! HC#)$=;&()wJS!*9!)*+,'(!2:U'*!%<!'(*59=!I8S!'*|2! (x; y) = (1;1) -!! ZV&4)$=;&F!h(YO5!%R%*!%*l!:4! x = y !'*+!Nm5!(5{5!2:;'!24!%<!2*i!2*€%!*59'!NZ'!*3)!'*+!v48V! ! y x + 2 − x y + 2 = 2(x 3 − y 3 ) ⇔ y 2 (x + 2)− x 2 ( y + 2) y x + 2 + x y + 2 = 2(x 3 − y 3 ) ⇔ (y − x) xy + 2(x + y) y x + 2 + x y + 2 + 2(x 2 + xy + y 2 ) ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ = 0 !-! b•'(!%R%*!.12!*3)!%*H'(!=5'*!*9!%*l!%<!'(*59=!.*5!_AS!I+,'(!2u!)*+,'(!2:U'*!2*H!*45-!X8S! '*5;'!%R%*!NO=!'OS!v‚!.*/'(!*598!T8{!'18!2u!)*+,'(!2:U'*!2*H!*45!%k4!*9!.*/'(!%*Y!)*Q)! v8S!:4!6578!.59'!_ASƒK-! Z[%)#;@)#<\&1)#])^) Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! „! Z[%):_)*KF!f5{5!*9!)*+,'(!2:U'*! x y + 4 − y x + 4 = 2(y − x) x 2 + y 2 − 2x + 2y = 8 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ -!qyvV! (x; y) = (−2;−2);(2;2) -!!! Z[%):_)*.F)f5{5!*9!)*+,'(!2:U'*! y x + 2 − x y + 2 = 2(x 3 − y 3 ) x 3 − 4x 2 + 4x −1 = 13− y 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ -!qyvV! (x; y) = (3;3) -! ndV!XU=!6+3%! y = x 6+4!P7!(5{5!)*+,'(!2:U'*V!! ! x 3 − 4x 2 + 4x −1 = 13− x 2 ⇔ (x 3 − 4x 2 + 4x − 3) + (2− 13− x 2 ) = 0 ⇔ (x − 3)(x 2 − x +1)+ x 2 − 9 13− x 2 + 2 = 0 ⇔ (x − 3)(x 2 − x +1+ x + 3 13− x 2 + 2 ) = 0 ⇔ x = 3 !-! IY! x 2 − x +1+ x + 3 13− x 2 + 2 > 0,∀x >−2 -!! HC#)$=;&(!n9!%<!'(*59=!I8S!'*|2! (x; y) = (3;3) -!! BI=)+)JKL*)>%M6NF!c*Y!_ASAW!NO!%R%!vj!2*€%!.*/'(!t=!2*Y{!=}'! x 2 + y 2 + z 2 = 2 -!XU=!(5R!2:…!'*†! '*|2!%k4!`5i8!2*H%! P = 1 x 2 + y 2 + 1 y 2 + z 2 + 1 z 2 + x 2 + 12 2− (x + y + z) 2 -! dY! x 2 + y 2 + z 2 = 2 ⇒ 2−(x + y + z) 2 = −2(xy + yz + zx ) -! p8S!:4V P = 1 x 2 + y 2 + 1 y 2 + z 2 + 1 z 2 + x 2 − 6 xy + yz + zx -! X4!%<V! ! 1 x 2 + y 2 ∑ = 1 2 x 2 + y 2 + z 2 x 2 + y 2 ∑ = 3 2 + 1 2 z 2 x 2 + y 2 ∑ -! X4!%<V! ! ( z 2 x 2 + y 2 ∑ )(x 2 y 2 + y 2 z 2 + z 2 x 2 ) ≥ x 4 + y 4 + z 4 -! wU!PJS!! 1 x 2 + y 2 ∑ ≥ 3 2 + 1 2 . x 4 + y 4 + z 4 x 2 y 2 + y 2 z 2 + z 2 x 2 = 1 2 + (x 2 + y 2 + z 2 ) 2 2(x 2 y 2 + y 2 z 2 + z 2 x 2 ) ≥ 1 2 + (x 2 + y 2 + z 2 ) 2 2(xy + yz + zx ) 2 = 1 2 + 2 (xy + yz + zx ) 2 ≥ 6 xy + yz + zx −4 -! d|8!`•'(!6\2!%*E'(!*\'! x = 0;x 2 + y 2 + z 2 = 2;xy + yz + zx = 1 3 -! wJS!(5R!2:…!'*†!'*|2!%k4!G!`•'(!?‡-!!!! b•'(!%R%*!2+,'(!2€!24!%<!.12!T8{!2ˆ'(!T8R2!v48V! 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&! B! 1 a 2 + b 2 + 1 b 2 + c 2 + 1 c 2 + a 2 + k a 2 + b 2 + c 2 ≥ α k ab + bc + ca -! X:Y'(!6<! ! α k = (2k + 9) / 2 for k ≤−4, α k = (k + 5) / 2 for −4 ≤ k ≤ 3, α k = 2 k +1 for k ≥ 3 !! Z[%)#;@)#<\&1)#]) Z[%):_)*KF!c*Y!4A`A%!NO!%R%!vj!2*€%!.*/'(!t=!2*Y{!=}'! ab + bc + ca =1 -!XU=!(5R!2:…!'*†!'*|2! %k4!`5i8!2*H%! P = 1 a 2 + b 2 + 1 b 2 + c 2 + 1 c 2 + a 2 + 1 a 2 + b 2 + c 2 -! Z[%):_)*.F!c*Y!4A`A%!NO!%R%!vj!2*€%!.*/'(!t=!2*Y{!=}'! a 2 + b 2 + c 2 = 2 -!XU=!(5R!2:…!'*†!'*|2! %k4!`5i8!2*H%! P = 1 a 2 + b 2 + 1 b 2 + c 2 + 1 c 2 + a 2 − 4 ab + bc + ca -! BI=)+)JKL*)>%M6NF!c*Y!4A`A%!NO!%R%!vj!2*€%!.*/'(!t=!%*H'(!=5'*!:•'(! ! 1 a 2 + b 2 + 1 b 2 + c 2 + 1 c 2 + a 2 + 8 a 2 + b 2 + c 2 ≥ 6 ab + bc + ca -! ! ! !! !! !!!! !! !! ! !!!!!!! . 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&)$U6)`U%()a*)@4b#c)24W&1)2d)#4_%)1%E&)1%E")>K) e%f&)4g)>0&1)23)24"I)489)Q)!"#$%&'()*+,-). = 1 x 2 + y 2 + 1 y 2 + z 2 + 1 z 2 + x 2 + 12 2− (x + y + z) 2 *! 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&! <! M!oV)LpB!)qrV!)estV)/uM)uV) !. 5 2 *!! qU%)#;@)#<v&1)#w)m!R1,1!P$AS3C!.T>3$! 2 2 x 2 −x − 2 x 2 −16.2 x 2 −x +16 = 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&! v! {€(|