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%)()\,]*.].*\^) 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 − (m + 2)x 2 + (2m +1)x + 2 (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";!5<1! m = −2 *!! =* !>?'!'!7@!:";!7A.!B/B!7A1!.A1!71@'! x 1 C!7A.!B/B!.1@D!.A1!71@'! x 2 (E%!B$%! x 1 2 − x 2 = −1 *!!!! Bi=).)j\c*)>%d6kF) E; F1,1!G$HI3J!.K?3$! log 18 (3 x + 2) = 1− x 2+ log 3 2 *! 0; F1,1!G$HI3J!.K?3$! sin 2x.sin x − 3π 2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ = 1+ sin x + tan x *!!! Bi=)7)j\c*)>%d6kF!>L3$!.LB$!G$M3! I = 2x − 1 x ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ln 2 x dx 1 2 ∫ *! Bi=)l)j\c*)>%d6kF!! E; >?'!B-B!()!.$/B!EC0CB!(E%!B$%!$E1!G$HI3J!.K?3$! az 2 + bz + c = 0;cz 2 + bz + a +16−16i = 0 :N3! O;!BP!!3J$1Q'!B$D3J!R&! 1+ 2i *! 0; #$%!3!R&!()!./!3$143!.$%,!'S3! C n 2 + A n 2 = n 2 + 35 *!>?'!()!$A3J!T$U3J!B$VE!W!.K%3J!T$E1! .K1@3! x 2 − 2 x 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ n *!!!! Bi=)^)j\c*)>%d6kF)#$%!$?3$!$XG!B$Y!3$Z.![\#]*[^\^#^]^!BP!.$@!.LB$!0_3J!"*!F`1!aCbCc!Rd3! RHe.!R&!.KD3J!71@'!B-B!7%A3!.$f3J![[^C#]C[^]*!>L3$!.$@!.LB$!T$)1!.V!g1Q3!\abc!5&!J1-!.K9! R<3!3$h.!BiE!T$%,3J!B-B$!J1YE!$E1!7Hj3J!.$f3J!#a!5&![^b*!! Bi=)-)j\c*)>%d6kF!>K%3J!T$U3J!J1E3!5<1!$Q!.KkB!.%A!7X!lWmO!B$%!0E!71@'![:"nono;C!\:onp=no;C! #:"np=nq;*!r12.!G$HI3J!.K?3$!'s.!G$f3J!71!tDE!0E!71@'![C\C#*!>?'!71@'!]!.K43!.1E!lO!(E%!B$%! .V!g1Q3![\#]!BP!.$@!.LB$!0_3J!=*!! Bi=),)j\c*)>%d6kF)>K%3J!'s.!G$f3J!5<1!.KkB!.%A!7X!lWm!B$%!$?3$!B$Y!3$Z.![\#]!BP! AC = 2BC G$HI3J!.K?3$!7Hj3J!B$u%![#!R&! 3x − y− 3 = 0 *!F`1!F!R&!.K`3J!.M'!BiE!.E'!J1-B! [#]!5&!71@'! H 3; 2 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ R&!.K/B!.M'!.E'!J1-B![\F*!r12.!G$HI3J!.K?3$!7Hj3J!.$f3J![]*! Bi=)a)j\c*)>%d6kF)F1,1!$Q!G$HI3J!.K?3$! x y +1 x +1 + y x +1 y +1 = x + y + 2xy 2 x 2 + y 2 −3xy + 5y = 4 x ( 5y −1 − x ) ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ (x, y ∈ !) *!!! Bi=)+)j\c*)>%d6kF!#$%!WCmCO!R&!B-B!()!.$/B!gHI3J!.$%,!'S3! x 2 + 8y 2 + 8z 2 = 20 *!>?'!J1-!.K9!R<3! 3$h.!BiE!01@D!.$VB! P = 2x 2 x 2 (4 − yz) + 8 + y + z x + y + z +1 − x 2 + ( y + z) 2 100 *! mmm!nLmmm) M!oV)LpB!)qrV!)estV)uv)/wM)wV) Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&)! =! Bi=)\)j.c*)>%d6kF!#$%!$&'!()! y = x 3 − (m + 2)x 2 + (2m +1)x + 2 (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";!5<1! m = −2 *!! =* !>?'!'!7@!:";!7A.!B/B!7A1!.A1!71@'! x 1 C!7A.!B/B!.1@D!.A1!71@'! x 2 (E%!B$%! x 1 2 − x 2 = −1 *!!!! "* v`B!(13$!./!J1,1*! =* >E!BPw! y ' = 3x 2 − 2(m + 2)x + 2m +1; y ' = 0 ⇔ x = 1 x = 2m +1 3 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ *! x@!:";!BP!=!B/B!.K9!T$1!m^!BP!=!3J$1Q'!G$M3!01Q.! ⇔ 2m +1 3 ≠ 1 ⇔ m ≠1 *! y!b2D! m >1 ⇒1< 2m +1 3 ⇒ x 1 = 1;x 2 = 2m +1 3 *! z4D!BdD!.K{!.$&3$w! 1 2 − 2m +1 3 = −1 ⇔ m = 5 2 (t / m) *! y!b2D! m <1 ⇒ 2m +1 3 <1 ⇒ x 1 = 2m +1 3 ,x 2 = 1 *! z4D!BdD!.K{!.$&3$w! 2m +1 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ 2 −1 = −1 ⇔ 2m +1 3 = 0 ⇔ m = − 1 2 (t / m) *! rZm! m = − 1 2 ;m = 5 2 R&!J1-!.K9!Bd3!.?'*!!!!! Bi=).)j\c*)>%d6kF) E; F1,1!G$HI3J!.K?3$! log 18 (3 x + 2) = 1− x 2+ log 3 2 *! 0; F1,1!G$HI3J!.K?3$! sin 2x.sin x − 3π 2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ = 1+ sin x + tan x *!!! E; c$HI3J!.K?3$!.HI3J!7HI3J!5<1w! log 18 (3 x + 2) = 1− x log 3 18 ⇔ log 18 (3 x + 2) = (1− x )log 18 3 ⇔ log 18 (3 x + 2) = log 18 3 1−x ⇔ 3 x + 2 = 3 1−x = 3 3 x ⇔ 3 2x + 2.3 x −3 = 0 ⇔ 3 x = 1 3 x = −3(l ) ⎡ ⎣ ⎢ ⎢ ⎢ ⇔ x = 0 *! rZm!G$HI3J!.K?3$!BP!3J$1Q'!gDm!3$h.! x = 0 *!! 0; x1|D!T1Q3w! cos x ≠ 0 ⇔ x ≠ π 2 + kπ,k ∈ ! *! c$HI3J!.K?3$!.HI3J!7HI3J!5<1w! 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&)! q! sin2x.cosx =1+sin x + sin x cos x ⇔ 2sin x.cos 3 x = cosx +sin x cos x + sin x ⇔ sin x cos x 2cos 2 x −1 ( ) = sin x + cos x ⇔ sin 2x cosx −sin x ( ) cos x +sin x ( ) = 2 sin x + cosx ( ) ⇔ sin x + cos x ( ) sin2x cos x −sin x ( ) − 2 ( ) = 0 ⇔ sin x + cos x = 0 (1) sin2x cos x −sin x ( ) − 2 = 0 (2) ⎡ ⎣ ⎢ ⎢ ⎢ *! y;!>E!BP! (1) ⇔ sin x = −cos x ⇔ tan x = −1 ⇔ x = − π 4 + kπ *! y;!>E!BP! VT (2) ≤ sin 2x . sin x − cos x − 2≤ 2 −2 < 0 :5U!3J$1Q';*! 5Zm!G$HI3J!.K?3$!BP!3J$1Q'!R&! x = − π 4 + kπ,k ∈ ! *!!!!! Bi=)7)j\c*)>%d6kF!>L3$!.LB$!G$M3! I = 2x − 1 x ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ln 2 x dx 1 2 ∫ *! >E!BP! I = 2 x ln 2 x 1 2 ∫ K ! "### $### − ln 2 x x dx 1 2 ∫ = 2K − ln 2 xd(ln x ) 1 2 ∫ = 2K − ln 3 x 3 2 1 = 2K − ln 3 2 3 *! xs.! u = ln 2 x dv = xdx ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ du = 2ln x x dx v = x 2 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ *! }Dm!KE! K = x 2 ln 2 x 2 2 1 − x ln x dx 1 2 ∫ M ! "#### $#### = 2ln 2 2−M *! xs.! u = ln x dv = xdx ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ du = dx x v = x 2 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⇒ M = x 2 ln x 2 2 1 − 1 2 x dx 1 2 ∫ = 2ln 2− 1 4 x 2 2 1 = 2ln 2− 3 4 *! r?!5Zm! I = 2 2ln 2 2−(2ln 2− 3 4 ) ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ − ln 3 2 3 = − ln 3 2 3 + 4 ln 2 2−4ln 2+ 3 2 *!!! Bi=)l)j\c*)>%d6kF!! B; >?'!B-B!()!.$/B!EC0CB!(E%!B$%!$E1!G$HI3J!.K?3$! az 2 + bz + c = 0;cz 2 + bz + a +16−16i = 0 :N3! O;!BP!!3J$1Q'!B$D3J!R&! 1+ 2i *! g; #$%!3!R&!()!./!3$143!.$%,!'S3! C n 2 + A n 2 = n 2 + 35 *!>?'!()!$A3J!T$U3J!B$VE!W!.K%3J!T$E1! .K1@3! x 2 − 2 x 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ n *!!!! E;!>$~%!J1,!.$12.!G$HI3J!.K?3$! az 2 + bz + c = 0 BP!3J$1Q'!:"y=1;!T$1!!! Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn! !"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A))) B4%)#%C#()DE#4$%&2:FG&)! •! ! a(1+ 2i ) 2 + b(1+ 2i) + c = 0 ⇔ −3a + b + c + (4a + 2b)i = 0 ⇔ −3a + b + c = 0 4a + 2b = 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ (1) *! >HI3J!./!G$HI3J!.K?3$! cz 2 + bz + a +16 −16i = 0 !BP!3J$1Q'!:"y=1;!T$1!! c(1+ 2i ) 2 + b(1+ 2i) + a +16−16i = 0 ⇔ c(−3+ 4i )+ b + 2bi + a +16−16i = 0 ⇔ (a + b − 3c +16) + 2(b + 2c −8)i = 0 ⇔ a + b − 3c +16 = 0 b + 2c − 8 = 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ (2) *! >€!:";!5&!:=;!(Dm!KE! (a;b;c) = (1;−2;5) *! ! 0;!>$~%!J1,!.$12.!.E!BPw! n(n −1) 2 + n(n −1) = n 2 + 35 ⇔ n 2 −3n −70 = 0 ⇔ n =10(t / m) n = −7(l ) ⎡ ⎣ ⎢ ⎢ *! +$1!7P! x 2 − 2 x 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ 10 = C 10 k x 20−5k (−2) k k=0 10 ∑ *! })!$A3J!T$U3J!B$VE!W!T$1! 20−5k = 0 ⇔ k = 4 *!!! rZm!()!$A3J!Bd3!.?'!R&! C 10 4 (−2) 4 *!! Bi=)^)j\c*)>%d6kF)#$%!$?3$!$XG!B$Y!3$Z.![\#]*[^\^#^]^!BP!.$@!.LB$!0_3J!"*!F`1!aCbCc!Rd3! RHe.!R&!.KD3J!71@'!B-B!7%A3!.$f3J![[^C#]C[^]*!>L3$!.$@!.LB$!T$)1!.V!g1Q3!\abc!5&!J1-!.K9! R<3!3$h.!BiE!T$%,3J!B-B$!J1YE!$E1!7Hj3J!.$f3J!#a!5&![^b*!! ! y;!•ZG!$Q!.KkB!.%A!7X!BP![:onono;C!\:Enono;C!#:En0no;C!]:on0no;C! [^:ononB;C!\^:EnonB;C!#^:En0nB;C]^:on0nB;*! +$1!7P! M (0;0; c 2 ),N ( a 2 ;b;0),P(0; b 2 ; c 2 ) *! >$~%!J1,!.$12.!.E!BPw! abc = 1 *!! y;!>E!BP! V BMNP = 1 6 BM ! "!! ,BN ! "!! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ .BP ! "! = 5abc 48 = 5 48 *!!!! y;!>L3$!7HeBw! d (CM ; A' N ) = CM ! "!! ,A' N ! "!!! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ .MA' ! "!!! CM ! "!! ,A' N ! "!!! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = abc 4b 2 c 2 + 4a 2 b 2 + 9a 2 c 2 = 1 4 a 2 + 4 c 2 + 9 b 2 *! }‚!gk3J!0h.!7f3J!.$VB![a!ƒFa!.E!BPw! ! 4 a 2 + 4 c 2 + 9 b 2 ≥ 3 4 a 2 . 4 c 2 . 9 b 2 3 = 3 144 3 ⇒ d (CM ; A' N ) ≤ 1 3 144 3 *! ]hD!0_3J!7A.!.A1! a 2 = c 2 = b 3 = 1 12 3 *!rZm!J1-!.K9!R<3!3$h.!BiE!T$%,3J!B-B$!J1YE!#a!5&![^b! 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&)! „! 0_3J! 1 3 144 3 *!! B4b)3(!\&1!.%-3!J1,!.$12.!$?3$!$XG!B$Y!3$Z.C!$?3$!RZG!G$HI3J!32D!T$P!.12G!BZ3!BP!.$@!J…3!.KkB! .%A!7X!7@!W‚!R†*! Bi=)-)j\c*)>%d6kF!>K%3J!T$U3J!J1E3!5<1!$Q!.KkB!.%A!7X!lWmO!B$%!0E!71@'![:"nono;C!\:onp=no;C! #:"np=nq;*!r12.!G$HI3J!.K?3$!'s.!G$f3J!71!tDE!0E!71@'![C\C#*!>?'!71@'!]!.K43!.1E!lO!(E%!B$%! .V!g1Q3![\#]!BP!.$@!.LB$!0_3J!=*!! >E!BPw! AB ! "!! = (−1;−2;0),AC ! "!! = (0;−2;3) ⇒ AB ! "!! ,AC ! "!! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = (−6;3;2) *! as.!G$f3J!71!tDE!0E!71@'![C\C#!3$Z3!! AB ! "!! ,AC ! "!! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = (−6;3;2) R&'!5uB!.I!G$-G!.Dm23!343!BP! G$HI3J!.K?3$!R&! −6(x −1)+ 3y + 2z = 0 ⇔ 6x − 3y − 2z −6 = 0 *! y;!F`1!]:ononE;!5<1!E‡o!.E!BP! S ABC = 1 2 AB ! "!! ,AC ! "!! ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 7 2 ;d (D;(ABC )) = −2a − 6 7 *! >E!BPw! V ABCD = 1 3 S ABC .d(D;(ABC )) = 1 3 . 7 2 . −2a − 6 7 = 2 ⇔ a = 3(t / m) a = −9(l ) ⎡ ⎣ ⎢ ⎢ *! rZm!71@'!Bd3!.?'!R&!]:ononq;*!!!!! Bi=),)j\c*)>%d6kF)>K%3J!'s.!G$f3J!5<1!.KkB!.%A!7X!lWm!B$%!$?3$!B$Y!3$Z.![\#]!BP! AC = 2BC G$HI3J!.K?3$!7Hj3J!B$u%![#!R&! 3x − y − 3 = 0 *!F`1!F!R&!.K`3J!.M'!BiE!.E'!J1-B! [#]!5&!71@'! H 3; 2 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ R&!.K/B!.M'!.E'!J1-B![\F*!r12.!G$HI3J!.K?3$!7Hj3J!.$f3J![]*! #$V3J!'13$!7HeB!v!.$DXB!7%A3!#]!5&! CH = 2 3 CD *!! >E!BP! AC = 2BC ⇒ CD = 3AD ⇒ ACD ! = 30 0 *! F`1!:En0;!R&!5uB!.I!G$-G!.Dm23!BiE!#]!.E!BPw! ! 3a −b 2 a 2 + b 2 = 3 2 ⇔ 3(a 2 + b 2 ) = ( 3a − b) 2 ⇔ 2b 2 + 2 3ab = 0 ⇔ b(b + 3a) = 0 ⇔ b = 0 b = − 3a ⎡ ⎣ ⎢ ⎢ ⎢ *! y;!b2D! b = 0 ⇒ CD : x − 3 = 0 *!! >%A!7X!71@'!#!R&!3J$1Q'!BiE!G$HI3J!.K?3$! x −3 = 0 3x − y − 3 = 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ x = 3 y = 2 3 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ C(3;2 3) *! >E!BP! CD ! "!! = 3 2 CH ! "!! = 3 2 0;− 4 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ = 0;−2 3 ( ) ⇒ D(3;0) *! xHj3J!.$f3J![]!71!tDE!]!5&!5DU3J!JPB!5<1!#]!BP!G$HI3J!.K?3$w! y = 0 *!!!! y;!b2D! b = − 3a ⇒ CD : x − 3y −1 = 0 *! >%A!7X!71@'!#!R&!3J$1Q'!BiE!$Q!G$HI3J!.K?3$! x − 3y −1= 0 3x − y − 3 = 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇔ x = 1 y = 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ C(1;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&)! 6! Ta!có! CD ! "!! = 3 2 CH ! "!! = 3; 3 ( ) ⇒ D(4; 3) .!! Đường!thẳng!AD!đi!qua!điểm!D!và!vuông!góc!với!CD!có!phương!trình: 3x + y −5 3 = 0 .! HC#)$=;&(!Vậy!có!hai!đường!thẳng!thoả!mãn!yêu!cầu!bài!toán!là 3x + y −5 3 = 0; y = 0 .!! BI=)J)KLM*)>%N6OF)Giải!hệ!phương!trình! x y +1 x +1 + y x +1 y +1 = x + y + 2xy 2 x 2 + y 2 −3xy + 5y = 4 x ( 5y −1 − x ) ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ (x, y ∈ !) .!!! Điều!kiện:! x ≥ 0; y ≥ 1 5 .!! Sử!dụng!bất!đẳng!thức!AM!–GM!ta!có:! x y +1 x +1 + y x +1 y +1 ≤ 1 2 x y +1 x +1 +1 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ + y x +1 y +1 +1 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = x + y + 2 2 x x +1 + y y +1 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ = x + y + 2 2 . x + y + 2xy (x +1)(y +1) = x + y + 2 2(xy + x + y +1) .(x + y + 2xy) .! Dấu!bằng!xảy!ra!kh!và!chỉ!khi! x = y .!! Kết!hợp!với!phương!trình!thứ!nhất!của!hệ!ta!có:! x + y + 2xy 2 ≤ x + y + 2 2(xy + x + y +1) .(x + y + 2xy) ⇔ x + y + 2 xy + x + y +1 ≥1 ⇔ xy ≤1 (1) .! Dấu!bằng!xảy!ra!khi!và!chỉ!khi! xy = 1 .!! Phương!trình!thứ!hai!của!hệ!tương!đương!với:! (x − y) 2 − xy + 5y = 4 x(5y −1) − 4x ⇔ (x − y) 2 − xy + 4x + 5y − 4 x(5y −1) = 0 ⇔ (x − y) 2 − xy +1+ 4x − 4 x(5y − 1) +5y −1 ( ) = 0 ⇔ (x − y) 2 − xy +1+(2 x − 5y −1) 2 = 0 ⇔ (x − y) 2 + (2 x − 5y −1) 2 = xy −1 .! Suy!ra! xy −1≥ 0 ⇔ xy ≥1 (2) .!! Từ!(1)!và!(2)!suy!ra! xy = 1 .!Vì!vậy! x = y xy =1 (x − y) 2 + (2 x − 5y −1) 2 = 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇔ x = 1 y = 1 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ (thử!lại!thấy!thoả! mãn).! Vậy!hệ!phương!trình!có!nghiệm!duy!nhất! (x; y) = (1;1) .!!!!!! BI=)+)KLM*)>%N6OF!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn! x 2 + 8y 2 + 8z 2 = 20 .!Tìm!giá!trị!lớn! nhất!của!biểu!thức! P = 2x 2 x 2 (4 − yz) + 8 + y + z x + y + z +1 − x 2 + ( y + z) 2 100 .! Lời$giải:$ 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&)! 7! Sử!dụng!bất!đẳng!thức!Cauchy!–!Schwarz!ta!có:! ! x 2 + ( y + z) 2 ≥ 1 2 x + y + z ( ) 2 .!! Sử!dụng!bất!đẳng!thức!AM!–GM!ta!có:! ! 2x 2 4x 2 + 8− x 2 yz = x 2x + 4 x − xyz 2 = x x + x + 4 x ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ − xyz 2 ≤ x x + 4 − xyz 2 .!! Sử!dụng!bất!đẳng!thức!Cauchy!–!Schwarz!ta!có:! ! x.yz + 2.y + 2.z + 2.1 ( ) 2 ≤ (x 2 +12)( y 2 z 2 + y 2 + z 2 +1) = x 2 +12 ( ) y 2 +1 ( ) z 2 +1 ( ) = 1 64 x 2 +12 ( ) 8y 2 + 8 ( ) 8z 2 + 8 ( ) ≤ 1 64 x 2 + 8y 2 + 8z 2 + 28 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ 3 = 64 ⇒ xyz + 2y + 2z ≤ 6 ⇒ x x + 4 − xyz 2 ≤ x x + y + z +1 .! Vì!vậy! P ≤ x + y + z x + y + z +1 − (x + y + z) 2 200 .!Đặt! t = x + y + z > 0 ta!có! P ≤ t t +1 − t 2 200 .! Xét!hàm!số! f (t) = t t +1 − t 2 200 !ta!có! f '(t) = 1 (t +1) 2 − t 100 ; f '(t) = 0 ⇔ t(t +1) 2 = 100 ⇔ t = 4 .! Vì!f’(t)!đổi!dấu!từ!dương!sang!âm!khi!đi!qua!t=4!nên! P ≤ f (t ) ≤ f (4) = 18 25 .! Dấu!bằng!đạt!tại! x = 2; y = z = 1 .!Vậy!giá!trị!lớn!nhất!của!P!bằng!18/25.!! BP94).() Sử dụng bất đẳng thức AM-GM cho hai số dương ta được: x 2 yz ≤ x 2 . y 2 + z 2 2 = 5 4 x 2 − x 4 16 ; 2xy + 2xz ≤ x 2 + 4 y 2 2 + x 2 + 4z 2 2 = x 2 + 2. 20− x 2 8 = 3 4 x 2 + 5; 2x ≤ x 2 2 + 2 ≤ ( x 4 16 +1) + 2 = x 4 16 + 3 Cộng theo vế ba bất đẳng thức cùng chiều ta thu được: x 2 yz + 2xy + 2xz + 2x ≤ 2x 2 + 8 . Suy ra: x 2 (4 − yz) + 8 = 2x(x + y + z +1)+ 2x 2 + 8−(x 2 yz + 2xy + 2xz + 2x ) ≥ 2x(x + y + z +1) . Mặt khác, theo bất đẳng thức Cauchy –Schwarz ta có: x 2 + ( y + z) 2 ≥ (x + y + z) 2 2 . Từ đó ta có được: P ≤ x + y + z x + y + z +1 − (x + y + z) 2 200 .!Ta!có!kết!quả!tương!tự!trên. 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&)! 8! Q4;&)RS#F)) +)!Mấu!chốt!bài!toán!tìm!được!điểm!rơi! x = 2; y = z = 1 .! +)!Đánh!giá! xyz + 2 y + 2z ≤ 6 .! TU%)#;@)#<V&1)#W)X!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn! x 2 + y 2 + z 2 = 26 .!Tìm!giá!trị!lớn! nhất!của!biểu!thức! P = xyz + 32 y + 45z .! Đ/S:! P max = P (1;3;4) = 288 .!!!!!!!!!! . 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%)(),]*.].*^) L4_%)1%E&)$U6)`U%()a*)@4b#c)24W&1)2d)#4_%)1%E&)1%E")>K) e%f&)4g)>0&1)23)24"I)489)Q)!"#$%&'()*+,-). + z +1 − x 2 + ( y + z) 2 100 *! mmm!nLmmm) M!oV)LpB!)qrV!)estV)uv)/wM)wV) 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&)! =! Bi=))j.c*)>%d6kF!#$%!$&'!()! . x1|D!T1Q3w! cos x ≠ 0 ⇔ x ≠ π 2 + kπ,k ∈ ! *! c$HI3J!.K?3$!.HI3J!7HI3J!5<1w! 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&)! q!