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Bj=),)k\d*)>%e6lF!NB%3A!'h.!>$i3A!5g1!.BlF!.%a!7m!noE!F$%!$C3$!.$%1!_`#q!FI!w1J3!.OF$!0K3A! 2 3 c!>$?@3A!.BC3$!7?k3A!F$x%!`q!R&! 3x − y = 0 *!yz3$!_!R&!$C3$!F$12D!5Df3A!AIF!FGH!`!.B43! 7?k3A!.$i3A! d : 3x + y = 0 *!NC'!.%a!7m!0)3!7z3$!_c`c#cq!012.!7z3$!_!FI!$%&3$!7m!w?@3A*! Bj=)`)k\d*)>%e6lF!L1,1!$J!>$?@3A!.BC3$! (x −1) 3 − 2x( y + 3) + 2y − 2 = 0 (x + y) 2 + y(1− −2x −1) = x 2 − (−2x −1) 3 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ (x, y ∈ ! ) *! Bj=)+)k\d*)>%e6lF)#$%!Hc0cF!R&!F-F!()!.$/F!w?@3A!.$%,!'{3! a 2 ( a c +1)+ b 2 ( b c +1) = 3 *!NC'!A1-!.B9! 3$|!3$M.!FGH!01eD!.$TF P = a + c b 3 + 2 + b + c a 3 + 2 − 2 a + b + c *! mmm!nLmmm) ) ) Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! <! ) PHÂN TÍCH BÌNH LUẬN ĐÁP ÁN Bj=)\)k.d*)>%e6lF!#$%!$&'!()! y = 3x +1 x −1 (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*! <* =12.!>$?@3A!.BC3$!.12>!.DE23!FGH!:";!012.!.12>!.DE23!FI!$J!()!AIF!0K3A! −1 *!! "* }QF!(13$!./!A1,1*! <* L1,!([!.12>!71e'! M (m; 3m +1 m −1 ),m ≠1 *!! }J!()!AIF!FGH!.12>!.DE23!.a1!b!R&! k = y'(m) = − 4 (m −1) 2 *! N$~%!A1,!.$12.!.H!FI!>$?@3A!.BC3$•!!! − 4 (m −1) 2 = −1 ⇔ (m −1) 2 = 4 ⇔ m = −1 m = 3 ⎡ ⎣ ⎢ ⎢ ⇒ M (−1;1) M (3;5) ⎡ ⎣ ⎢ ⎢ *! €;!=g1!b:s"r";!.H!FI!.12>!.DE23! y = −x *! €;!=g1!b:trX;!.H!FI!.12>!.DE23! y = −x + 8 *! HC#)$=;&(!=VE!FI!$H1!.12>!.DE23!FS3!.C'!R&! y = −x; y = −x + 8 *!!!!! Bj=).)k\d*)>%e6lF)) H; L1,1!>$?@3A!.BC3$! cos3x − 4 cos2x + 8 cos x − 5 = 0 *!! 0; L1,1!0M.!>$?@3A!.BC3$! log 2 x + 2log 2x 2 ≥ 2 *!! H; u$?@3A!.BC3$!.?@3A!7?@3A!5g1•!! ! 4cos 3 x −3cosx −4(2cos 2 x −1) + 8cos x −5 = 0 ⇔ 4cos 3 x −8cos 2 x + 5cos x −1= 0 ⇔ (cosx −1)(2cosx −1) 2 = 0 ⇔ cos x =1 cos x = 1 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⇔ x = k2π x = ± π 3 + k2π ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ *! HC#)$=;&(!=VE!>$?@3A!.BC3$!FI!3A$1J'!R&! x = k2π, x = ± π 3 + k2π,k ∈ ! *!!! #PD!<•! cos2x =1+ sin2x ⇔ cos2x −sin2x =1 ⇔ cos(2x + π 4 ) = 1 2 ⇔ 2x + π 4 = π 4 + k2π 2x + π 4 = − π 4 + k2π ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⇔ x = kπ x = − π 4 + kπ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ,k ∈ ! *! =VE!>$?@3A!.BC3$!FI!3A$1J'!R&! x = kπ, x = − π 4 + kπ,k ∈ ! *!! 0; y1]D!j1J3•! x > 0 2x ≠1 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ 0 < x ≠ 1 2 *! `M.!>$?@3A!.BC3$!.?@3A!7?@3A!5g1•! log 2 x + 2 log 2 2x ≥ 2 ⇔ log 2 x + 2 log 2 x +1 ≥ 2 *! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! t! yh.! t = log 2 x c!0M.!>$?@3A!.BC3$!.B•!.$&3$•! ! t + 2 t +1 ≥ 2 ⇔ t 2 + t + 2 t +1 ≥ 2 ⇔ t 2 −t t +1 ≥ 0 ⇔ t ≥1 −1< t ≤ 0 ⎡ ⎣ ⎢ ⎢ ⇔ log 2 x ≥1 −1< log 2 x ≤ 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⇔ x ≥ 2 1 2 < x ≤1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ *! HC#)$=;&(!NV>!3A$1J'!FGH!0M.!>$?@3A!.BC3$!R&! S = 1 2 ;1 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎤ ⎦ ⎥ ⎥ ∪ 2;+∞ ⎡ ⎣ ⎢ ) *!!! !!!! F; y1]D!j1J3•! x >1 x < −2 ⎡ ⎣ ⎢ ⎢ *! `M.!>$?@3A!.BC3$!.?@3A!7?@3A!5g1•! ! x −1 x > x −1 x + 2 ⇔ (x −1)( 1 x − 1 x + 2 ) > 0 ⇔ 2(x −1) x (x + 2) > 0 ⇔ x >1 x < −2 ⎡ ⎣ ⎢ ⎢ *! =VE!.V>!3A$1J'!FGH!0M.!>$?@3A!.BC3$!R&! S = (−∞;−2) ∪ (1;+∞) *! Bj=)7)k\d*)>%e6lF!NO3$!.OF$!>$P3! I = e x e x + 2 e x + 3 dx 0 ln 6 ∫ *! yh.! t = e x + 3 ⇒ e x = t 2 − 3 ⇒ e x dx = 2tdt *! NH!FI•! ! I = 2tdt t 2 −3+ 2t 2 3 ∫ = 2tdt (t −1)(t + 3) 2 3 ∫ = 1 2 3 t + 3 + 1 t −1 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 3 ∫ dt = 1 2 3ln t + 3 + ln t −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 3 2 = 1 2 (3ln 6 5 + ln 2) *! Bj=)])k\d*)>%e6lF))) H; LQ1! z 1 ,z 2 R&!$H1!3A$1J'!FGH!>$?@3A!.BC3$! z 2 − 3z + 5 = 0 *!NC'!>$S3!,%!FGH!()!>$TF! w = (z 1 2 + z 2 2 ).i + 2−3i *!) 0; LQ1!U!R&!.V>!$W>!F-F!()!./!3$143!FI!X!F$Y!()!5&!F$1H!$2.!F$%!Z*!NO3$!()!>$S3!.[!FGH!U!5&! .\3A!FGH!.M.!F,!F-F!>$S3!.[!7I*) H; NH!FI•! Δ = 3 2 − 4.5 = −11= 11i 2 ⇒ z 1 = 3− 11i 2 ,z 2 = 3+ 11i 2 *! +$1!7I•! w = ( 3− 11i 2 ) 2 + ( 3+ 11i 2 ) 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ .i + 2−3i = 2− 4i *! Kết$luận:!=VE!>$S3!,%!FGH!‚!0K3A!sƒ*!! B4c)3F! z 1 2 + z 2 2 = (z 1 + z 2 ) 2 − 2z 1 z 2 = 3 2 − 2.5 = −1 *!! 0; #-F!F$Y!()!.$DmF!U!R&! 10003,10010, ,99995 *!#$„3A!RV>!.$&3$!'m.!FM>!()!Fm3A!5g1!()!$a3A! 7SD!0K3A!"………tc!Ff3A!(H1!0K3A!Zc!()!$a3A!FD)1!R&!††††X*!!!! =VE!.B%3A!U!FI!.M.!F,!3!>$S3!.[!.$%,!'{3• 99995 =10003+ (n −1).7 ⇔ n =12857 *! 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&! ƒ! €;!N\3A!FGH!.M.!F,!F-F!>$S3!.[!FGH!U!0K3A• S = (a 1 + a n ).n 2 = (10003+ 99995).12857 2 = 707122143 *!! #$„!‡!.\3A!FGH!3!()!$a3A!7SD!FGH!FM>!()!Fm3A! S = (a 1 + a 2 ).n 2 *! oU%)#;@)#<p&1)#q)m) LQ1!U!R&!.V>!$W>!.M.!F,!F-F!()!./!3$143!A8'!ƒ!F$Y!()!5&!F$1H!$2.!F$%!ˆ*!NO3$!()!>$S3!.[!FGH!U! 5&!.\3A!FGH!.M.!F,!F-F!>$S3!.[!7I*!! Bj=)_)k\d*)>%e6lF!#$%!$C3$!F$I>!.H'!A1-F!7]D!^*_`#!FI!7-E!_`#!R&!.H'!A1-F!7]D!Fa3$!<H*!LQ1! bcd!RS3!R?W.!R&!.BD3A!71e'!FGH!F-F!7%a3!`#c^b*!`12.!_d!5Df3A!AIF!5g1!'h.!>$i3A!:^`#;*! NO3$!.$e!.OF$!j$)1!F$I>!^*_`#!5&!j$%,3A!F-F$!A1YH!$H1!7?k3A!.$i3A!^_!5&!`d*! ! €;!LQ1!n!R&!.BQ3A!.P'!.H'!A1-F!_`#c!.$~%!A1,!.$12.!.H!FI•! SO ⊥ (ABC ) *!! €;!NH!FI•! AM = AB.sin60 0 = 2a. 3 2 = a 3;SO = 2 3 AM = 2a 3 3 *! q%! AN ⊥ (SBC ) ⇒ AN ⊥ SM *!=C!_d!5‰H!R&!7?k3A!FH%!5‰H!R&! .BD3A!.DE23!FGH!.H'!A1-F!^_b!343!^_b!FP3!.a1!_*! ^DE!BH• SA = AM = a 3 *!! €;!NH'!A1-F!5Df3A!_^n!FI! SO = SA 2 − AO 2 = 3a 2 − 12a 2 9 = a 15 3 *! =C!5VE! V S .ABC = 1 3 SO.S ABC = 1 3 . a 15 3 . (a 3) 2 . 3 4 = a 3 5 4 :75 ;*! €;!NO3$!w:^_r`d;*! #-F$!"•!^[!wl3A!$J!.BlF!.%a!7m!noEp!FI!A)F!b:…r…r…;c!`:sHr…r…;c!#:Hr…r…;c! O(0; a 3 3 ;0),A(0;a 3;0),S (0; a 3 3 ; 2a 3 3 ) *! NH!FI!d!R&!.BD3A!71e'!^b!343! N (0; a 3 6 ; a 3 3 ) *! NH!FI•!!! SA ! "! = (0; 2a 3 3 ;− 2a 3 3 ),BN ! "!! = (a; a 3 6 ; a 3 3 ),BA ! "! = (a;a 3;0) ⇒ SA ! "! ,BN ! "!! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = (a 2 ;− 2a 2 3 3 ;− 2a 2 3 3 ), SA ! "! ,BN ! "!! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ .BA ! "! = −a 3 *! =C!5VE!! ! d (SA,BN ) = SA ! "! ,BN ! "!! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ .BA ! "! SA ! "! ,BN ! "!! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = a 3 a 4 + 4a 4 3 + 4a 4 3 = a 33 11 !*! +2.!RDV3•! V S .ABC = a 3 5 4 ;d (SA;BN ) = a 33 11 *!! BI94).(!}C3$!j$f3A!A1H3!.$DS3!.D‡•! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! X! NH!FI•! AN ⊥ (SBC ) 343!R„F!3&E!w/3A!Ra1!$C3$!F%1!7-E!R&!^`#!5&!7?k3A!FH%!R&!_d•! +Š!.1H!^o‹‹`d!.H!FI!`d‹‹:^_c^o;!5C!5VE! ! d (BN ;SA) = d (BN ;(SA,Sx )) = d(N ;(SA,Sx )) *! U~'!FV>!3$V.!F$1!.12.!.B%3A!`%o!7]!()!"ƒ*!!!! Bj=)-)k\d*)>%e6lF!NB%3A!j$f3A!A1H3!5g1!$J!.BlF!.%a!7m!noEp!F$%!$C3$!0C3$!$&3$!_`#q!FI! _:"r<rst;c!`:trXrsZ;c!#:"rtrs<;*!NC'!.%a!7m!71e'!q*!=12.!>$?@3A!.BC3$!'h.!>$i3A!:u;!71!vDH!q!5&! F$TH!7?k3A!.$i3A!n`!5g1!n!R&!A)F!.%a!7m*! €;!q%!_`#q!R&!$C3$!0C3$!$&3$!343! AB ! "!! = DC ! "!! = (2;3;−4) ⇒ 1− x D = 2 3− y D = 3 −2− z D = −4 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇔ x D = −1 y D = 0 z D = 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇒ D(−1;0;2) *! €;!NH!FI•! OB ! "! = (3;5;−7),OD ! "!! = (−1;0;2) *! bh.!>$i3A!:u;!71!vDH!q!5&!F$TH!7?k3A!.$i3A!n`!343!FI!5.>.!R& n ! = OB " !" ,OD " !"" ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = (10;1;5) *! =C!5VE! (P ) :10(x +1) +1(y −0) + 5(z −2) = 0 ⇔ (P ) :10x + y + 5z = 0 *! HC#)$=;&(!=VE! D(−1;0;2),(P ) :10x + y + 5z = 0 *!! Bj=),)k\d*)>%e6lF!NB%3A!'h.!>$i3A!5g1!.BlF!.%a!7m!noE!F$%!$C3$!.$%1!_`#q!FI!w1J3!.OF$!0K3A! 2 3 c!>$?@3A!.BC3$!7?k3A!F$x%!`q!R&! 3x − y = 0 *!yz3$!_!R&!$C3$!F$12D!5Df3A!AIF!FGH!`!.B43! 7?k3A!.$i3A! d : 3x + y = 0 *!NC'!.%a!7m!0)3!7z3$!_c`c#cq!012.!!7z3$!_!FI!$%&3$!7m!w?@3A*! ! }H1!7?k3A!.$i3A!`q!5&!w!FŒ.!3$HD!.a1!A)F!.%a!7m!n:…r…;!5&!AIF!A1YH! F$„3A!0K3A! ! cos(d 1 ,d 2 ) = 3. 3 −1.1 3+1. 3+1 = 1 2 ⇒ BOA ! = 60 0 *! €;!LQ1!•!R&!.P'!$C3$!.$%1!_`#qc!.$C!.H'!A1-F!n_#!5Df3A!.a1!_!343! IAB ! = BOA ! = 60 0 *!! NH!FI•! S ABCD = 4S IAB = 2AI .AB sin 60 0 = 3 OA.sin 60 0 ( ) . OA.tan 60 0 ( ) = 3 3 2 OA 2 *! N$~%!A1,!.$12.!.H!FI• 3 3 2 OA 2 = 2 3 ⇔ OA 2 = 4 3 *!! LQ1!_:orE;!5g1!oŽ…!.H!FI!$J!>$?@3A!.BC3$•! x 2 + y 2 = 4 3 3x + y = 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇒ A( 1 3 ;−1) *! €;!y?k3A!.$i3A!_`!71!vDH!_!5&!5Df3A!AIF!5g1!w!343!FI!>.!R&! x − 3y− 4 3 = 0 *! N%a!7m!71e'!`!R&!3A$1J'!FGH!$J! x − 3y− 4 3 = 0 3x − y = 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇔ x = − 2 3 y = −2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇒ B(− 2 3 ;−2) *! 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! €;!y?k3A!.$i3A!_•!5Df3A!AIF!5g1!`q!FI!>.!R&! x + 3y + 2 3 = 0 *! N%a!7m!71e'!•!R&!3A$1J'!FGH!$J! 3x − y = 0 x + 3y + 2 3 = 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇒ I − 1 2 3 ;− 1 2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ *! €;!=C!•!R&!.BD3A!71e'!FGH!_#!5&!`q!343! C − 2 3 ;0 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ,D 1 3 ;1 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ *! HC#)$=;&(!=VE! A( 1 3 ;−1),B (− 2 3 ;−2),C (− 2 3 ;0),D( 1 3 ;1) *!!!!!! Bj=)`)k\d*)>%e6lF!L1,1!$J!>$?@3A!.BC3$! (x −1) 3 − 2x( y + 3) + 2y − 2 = 0 (x + y) 2 + y(1− −2x −1) = x 2 − (−2x −1) 3 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ *! y1]D!j1J3•! x ≤− 1 2 *!u$?@3A!.BC3$!.$T!$H1!FGH!$J!.?@3A!7?@3A!5g1•! ! y 2 + 2xy + y + (−2x −1) 3 = y −2x −1 ⇔ y 2 − y −2x −1 + (2x +1)y −(2x +1) −2x −1 = 0 ⇔ (y − −2x −1)( y + 2x +1) = 0 ⇔ y = −2x −1 y = −2x −1 ⎡ ⎣ ⎢ ⎢ ⎢ *! €;!d2D! y = −2x −1 .$HE!5&%!>$?@3A!.BC3$!7SD!FGH!$J!.H!7?WF•! x 3 + x 2 −5x −5 = 0 ⇔ (x +1)(x 2 −5) = 0 ⇔ x = −1(t / m) x = − 5(t / m) x = 5(l ) ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⇒ x = −1, y =1 x = − 5, y = 2 5 −1 ⎡ ⎣ ⎢ ⎢ ⎢ !*! €;!d2D! y = −2x −1 .$HE!5&%!>$?@3A!.BC3$!7SD!FGH!$J!.H!7?WF•! ! (x −1) 3 + ( −2x −1 +1) 3 = 0 ⇔ −2x −1 +1 = 1− x ⇔ −2x −1 = −x ⇔ x ≤ 0 x 2 = −2x −1 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ x = −1 ⇒ y = 1 *! =VE!$J!>$?@3A!.BC3$!FI!$H1!3A$1J'!R&! (x; y) = (−1;1);(− 5;2 5 −1) *! Bj=)+)k\d*)>%e6lF)#$%!Hc0cF!R&!F-F!()!.$/F!w?@3A!.$%,!'{3! a 2 ( a c +1)+ b 2 ( b c +1) = 3 *!NC'!A1-!.B9! 3$|!3$M.!FGH!01eD!.$TF P = a + c b 3 + 2 + b + c a 3 + 2 − 2 a + b + c *! ^[!wl3A!0M.!7i3A!.$TF!_b!•Lb:wa3A!3A?WF!wMD;!.H!FI•! ! a + c b 3 + 2 = 1 2 a + c − (a + c)b 3 b 3 + 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ≥ 1 2 a + c − (a + c)b 3 3b ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 1 2 a + c − (a + c)b 2 3 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ *! N?@3A!./!.H!FI•! b + c a 3 + 2 ≥ 1 2 b + c − (b + c )a 2 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&! Z! ^DE!BH•! ! a + c b 3 + 2 + b + c a 3 + 2 ≥ 1 2 a + b + 2c − (a + c)b 2 + (b + c )a 2 3 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 1 2 a + b + 2c − ab(a + b) + c (a 2 + b 2 ) 3 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ≥ 1 2 (a + b + c ) *! `•1!5C!.$~%!A1,!.$12.!.H!FI•! c(3−a 2 −b 2 ) = a 3 + b 3 ≥ ab(a + b) ⇒ ab(a + b) + c (a 2 + b 2 ) ≤ 3c *!! =C!5VE!! ! P ≥ 1 2 (a + b + c )− 2 a + b + c = ( a + b + c −2) 2 2 − 2 ≥−2 *! qMD!0K3A!7a.!.a1! a + b + c = 2 a = b =1 a 2 (a + c) + b 2 (b + c ) = 3c ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⇔ a = b =1 c = 2 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ *! =VE!A1-!.B9!3$|!3$M.!FGH!u!0K3A!s<*!!! oU%)#;@)#<p&1)#q)m)) oU%):O)*\F!#$%!Hc0cF!R&!F-F!()!.$/F!w?@3A!F$T3A!'13$!BK3A! ! a + b c 3 + 2 + b + c a 3 + 2 + c + a b 3 + 2 ≥ a + b + c − a 3 + b 3 + c 3 3 *! oU%):O)*.F!#$%!Hc0cF!R&!F-F!()!.$/F!w?@3A!.$%,!'{3! a 3 + b 3 + c 3 = 3 *!NC'!A1-!.B9!3$|!3$M.!FGH! 01eD!.$TF! P = a + b c 3 + 2 + b + c a 3 + 2 + c + a b 3 + 2 − 2(a 2 + b 2 + c 2 ) *! oU%):O)*7F!#$%!Hc0cF!R&!F-F!()!.$/F!w?@3A!.$%,!'{3! a 2 ( a c +1)+ b 2 ( b c +1) = 3 *!NC'!A1-!.B9!3$|! 3$M.!FGH!01eD!.$TF! P = (a + b + c ) 2 b + c a 3 + 2 + a + c b 3 + 2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ − 9(a + b + c) 2 *!!!!!! ! ! . Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! "! H4"I)1%J%)>K)L!ML)N=O9)P%E)Q)L4RS()/T&1)L4U&4)VE6) DW&()L"I&X)/Y)Z[)]^_*) V1US)#4%)()*`^*7^.*\_) L4a%)1%E&)$U6)bU%()`*)@4c#d)24W&1)2e)#4a%)1%E&)1%E")>K) f%g&)4h)>0&1)23)24"I)489)Q)!"#$%&'()*+,-). P = a + c b 3 + 2 + b + c a 3 + 2 − 2 a + b + c *! mmm!nLmmm) ) ) Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! <! ) PHÂN. ≠ 1 2 *! `M.!>$?@3A!.BC3$!.?@3A!7?@3A!5g1•! log 2 x + 2 log 2 2x ≥ 2 ⇔ log 2 x + 2 log 2 x +1 ≥ 2 *! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&! t! yh.!