!"#$%&'('%)*%+",+% /0%&'1%+"23%)45&%+"65"%517% "89:;<=>%?@AB%CBB%C?C%% )D<E%FG%<HIJ%K%HLM%N;<H%<HO<%PQ%RS;%HLM%THU%%% 0H;%9;V9>%7W9H:;<FNXY<! "! !H8Z%E;[;%R\%+",+%-Q]M%&;W%^%+H_`>%)a<E%+Hb<H%5WJ% 7c<>%+8Z<d%)*%e/%CKfg?% 5Eb`%9H;%>%h?f?ifC?hg% +Hj;%E;W<%:bJ%kb;>%hl?%THm9n%FHc<E%Fo%9Hj;%E;W<%E;W8%R\% p;q<%Hr%RD<E%FG%FH8Z%HLM%^%"89:;<=>%?@AB%CBB%C?C%^%0H;%9;V9>%sssXJW9H:;<FNXY<% 0tQ%h%uCn?%R;oJvX%#$%!$&'!()! y = 2x x− 2 (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*! <* #$%!71='!>:<?<;*!@A'!71='!B!.$CDE!:";!(F%!E$%!.12G!.12G!.CH23!EIF!:";!.J1!B!5CK3L!LME!5N1! 7OP3L!.$Q3L!>B*! 0tQ%C%uhn?%R;oJvX% F; R1,1!G$OS3L!.TA3$! tanx −1= 2 sin(x + 3π 4 ) *!! 0; #$%!()!G$UE!V!.$%,!'W3! (1+ 2i).z = 3+ i *!@X3$!L1-!.T9!EIF!01=C!.$UE! A = z 4 − z 2 +1 *! 0tQ%K%u?ng%R;oJvX!R1,1!G$OS3L!.TA3$! log 2 x +1 x = log 4 x 2 +1 *! 0tQ%i%uhn?%R;oJvX!R1,1!0Y.!G$OS3L!.TA3$! x +1+ x 2 + x +1 − x 2 − x +1+1 x ≤ 1 x − x −1 *! 0tQ%g%uhn?%R;oJvX!@X3$!.XE$!G$Z3! I = (x +1) 2 − x.ln x x 2 dx 2 3 ∫ *!! 0tQ%B%uhn?%R;oJvX!#$%!$A3$![\3L!.T]!7U3L!^_#*^`_`#`!EM!7-H!^_#![&!.F'!L1-E!7aC!EJ3$!Fb! A 'B = a 7 2 *!Rc1!Bbdbe![f3![Og.![&!.TC3L!71='!E-E!EJ3$!^`_`b^`#`b##`*!@X3$!.$=!.XE$!h$)1![\3L! .T]!^_#*^`_`#`!5&!h$%,3L!E-E$!.i!71='!^!723!'j.!G$Q3L!:Bde;*!! 0tQ%A%uhn?%R;oJvX!@T%3L!'j.!G$Q3L!5N1!.T]E!.%J!7D!klH!E$%!7OP3L!.Tm3! (C ) : x 2 + y 2 = 1 *!Rc1! ^![&!71='!.$CDE!7OP3L!.$Q3L! y −3 = 0 !5&!_b#![f3![Og.![&!E-E!.12G!71='!EIF!.12G!.CH23!hn!.i!^! 723!:#;*!oCF!_!hn!7OP3L!.$Q3L!(%3L!(%3L!5N1!^#!Ep.!:#;!.J1!qb!^q!Ep.!:#;!.J1!r*!@A'!.%J!7D! 71='!^b!012.!_r!Ep.!^#!.J1!71='!>:"?<;*! 0tQ%l%uhn?%R;oJvX%@T%3L!h$K3L!L1F3!5N1!$s!.T]E!.%J!7D!klHV!E$%!$F1!71='!^:"?<?t";!5&!_:u?u?<;*! @A'!.%J!7D!71='!>!.$%,!'W3! IA !"! = 2IB !"! *!v12.!G$OS3L!.TA3$!7OP3L!.$Q3L! Δ 71!wCF!>b!Ep.!.T]E!kV! 5&!5CK3L!LME!5N1!7OP3L!.$Q3L!^_*!!! 0tQ%@%u?ng%R;oJvX%@i!'D.!3LZ3!$&3L!<u!EZC!$x1b!.T%3L!7M!EM!y!EZC!$x1!h$M!3LOP1!.F!lZH!z/3L! .$&3$!$F1!7a!.$1!'{1!7a!.$1!L8'!"u!EZCb!5&!E-E!EZC!.T%3L!'D.!7a!7OgE!7-3$!()!.$U!./!.i!#ZC!"! 723!#ZC!"u*!@X3$!l-E!(CY.!7=!lZH!z/3L!7OgE!$F1!7a!.$1!'&!'{1!7a!.$1!7aC!L8'!<!EZC!$x1!h$M*!! 0tQ%h?%uhn?%R;oJvX!#$%!Fb0bEb'b3bG![&!E-E!()!.$/E!.$%,!'W3! a 2 + b 2 + c 2 = m 2 + n 2 + p 2 = 9 *!@A'! L1-!.T9![N3!3$Y.!EIF!01=C!.$UE! P = 9− a− 2b−2c + 9− m− 2n− 2p + 9− am− bn− cp *!! www"x+www% !"#$%&'('%)*%+",+% /0%&'1%+"23%)45&%+"65"%517% "89:;<=>%?@AB%CBB%C?C%% )D<E%FG%<HIJ%K%HLM%N;<H%<HO<%PQ%RS;%HLM%THU%%% 0H;%9;V9>%7W9H:;<FNXY<! <! % ,"y5%+z0"%{|5"%p.}5%)$,%$5% 0tQ%h%uCn?%R;oJvX%#$%!$&'!()! y = 2x x− 2 (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*! <* #$%!71='!>:<?<;*!@A'!71='!B!.$CDE!:";!(F%!E$%!.12G!.12G!.CH23!EIF!:";!.J1!B!5CK3L!LME!5N1! 7OP3L!.$Q3L!>B*! "* |cE!(13$!./!L1,1*! <* Rc1! M(m; 2m m−2 )∈ (1),m ≠ 2 b!(CH!TF!$s!()!LME!EIF!7OP3L!.$Q3L!>B![&! k IM = y M − y I x M − x I = 2m m−2 − 2 m−2 = 4 (m− 2) 2 *! };!|s!()!LME!EIF!.12G!.CH23!5N1!:";!.J1!B![&! k = y'(m) = −4 (m− 2) 2 *! @$~%!L1,!.$12.!.F!EM•! k IM .k = −1⇔ −16 (m− 2) 4 = −1⇔ (m −2) 4 = 16 ⇔ m−2 = 2 m−2 = −2 ⎡ ⎣ ⎢ ⎢ ⎢ ⇔ m = 4 m = 0 ⎡ ⎣ ⎢ ⎢ ⎢ *! };!v€H!EM!$F1!71='!B:y?y;!5&!B:u?u;*!!!!! 0tQ%C%uhn?%R;oJvX% F; R1,1!G$OS3L!.TA3$! tanx −1= 2 sin(x + 3π 4 ) *!! 0; #$%!()!G$UE!V!.$%,!'W3! (1+ 2i).z = 3+ i *!@X3$!L1-!.T9!EIF!01=C!.$UE! A = z 4 − z 2 +1 *! F; •1aC!h1s3•! cosx ≠ 0 ⇔ x ≠ π 2 + kπ,k ∈ ! *! e$OS3L!.TA3$!.OS3L!7OS3L!5N1•! sin x− cosx cosx = −sin x + cosx ⇔ (sin x−cosx)(cosx +1) = 0 ⇔ sin x = cosx cosx = −1 ⎡ ⎣ ⎢ ⎢ ⎢ ⇔ tanx = 1 cosx = −1 ⎡ ⎣ ⎢ ⎢ ⎢ ⇔ x = π 4 + kπ x = π + k2π ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ,k ∈ ! *! v€H!G$OS3L!.TA3$!EM!3L$1s'! x = π 4 + kπ;x = π + k2π,k ∈ ! *!! 0; !@F!EM•! z = 3+ i 1+ 2i = (3+ i)(1− 2i) 5 = 5−5i 5 = 1− i ⇒ z = 2 *! q%!7M! A = ( 2) 4 − ( 2) 2 +1= 4− 2+1= 3 *!! 0tQ%K%u?ng%R;oJvX!R1,1!G$OS3L!.TA3$! log 2 ( x +1 x )= log 4 x 2 +1 *! !"#$%&'('%)*%+",+% /0%&'1%+"23%)45&%+"65"%517% "89:;<=>%?@AB%CBB%C?C%% )D<E%FG%<HIJ%K%HLM%N;<H%<HO<%PQ%RS;%HLM%THU%%% 0H;%9;V9>%7W9H:;<FNXY<! ‚! •1aC!h1s3•! x ≠ 0 x +1 x > 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇔ x > 0 x <−1 ⎡ ⎣ ⎢ ⎢ ⎢ *!e$OS3L!.TA3$!.OS3L!7OS3L!5N1•! log 2 ( x +1 x )= log 2 x + log 2 2 ⇔ log 2 ( x +1 x )= log 2 2 x ⇔ x +1 x = 2 x ⇔ x > 0 2x = x +1 x ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ x <−1 −2x = x +1 x ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⇔ x = 1 *! v€H!G$OS3L!.TA3$!EM!3L$1s'!zCH!3$Y.! x = 1 *!! 0tQ%i%uhn?%R;oJvX!R1,1!0Y.!G$OS3L!.TA3$! x +1+ x 2 + x +1 − x 2 − x +1+1 x ≤ 1 x − x −1 *! •1aC!h1s3•! x > 0 *! _Y.!G$OS3L!.TA3$!.OS3L!7OS3L!5N1•!!! x +1+ x 2 + x +1 − x 2 − x +1+1 x ≤ 1 x − x −1 (x +1)+ (x +1) 2 −(x +1)+1 + x +1≤ 1 x + 1 x 2 − 1 x +1 + 1 x (1) *! ƒ„.!$&'!()! f (t)= t+ t 2 − t +1 + t b!.F!EM•! f '(t) = 1+ 1+ 2t −1 2 t 2 − t +1 2 t + t 2 − t +1 = 1+ 2t −1+ 2 t 2 − t +1 4 t + t 2 − t +1. t 2 − t +1 > 0 b!! 0…1!5A! 2t −1+ 2 t 2 −t +1 = 4t 2 − 4t + 4 + 2t −1= (2t −1) 2 + 3 + 2t −1> 2t −1 + 2t −1≥ 0 *!!!! q%!7M!†:.;![&!$&'!783L!0123*!vA!5€H (1) ⇔ x +1≤ 1 x ⇔ x 2 + x −1 x ≤ 0 ⇔ 0< x ≤ −1+ 5 2 (do x > 0) *!! v€H!.€G!3L$1s'!EIF!0Y.!G$OS3L!.TA3$![&! S = 0; −1+ 5 2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎤ ⎦ ⎥ ⎥ ⎥ *!! {b;%9OT%9P~<E%9•w%% R1,1!0Y.!G$OS3L!.TA3$! x +1+ x 2 + x +1 − 2x + 4x 2 − 2x +1 ≥ x −1 *!•‡(•! S = 1;+∞ ⎡ ⎣ ⎢ ) *!!! 0tQ%g%uhn?%R;oJvX!@X3$!.XE$!G$Z3! I = (x +1) 2 − x.ln x x 2 dx 2 3 ∫ *!! };!@F!EM•!! I = (x +1) 2 x 2 dx 2 3 ∫ − ln x x dx 2 3 ∫ *! !"#$%&'('%)*%+",+% /0%&'1%+"23%)45&%+"65"%517% "89:;<=>%?@AB%CBB%C?C%% )D<E%FG%<HIJ%K%HLM%N;<H%<HO<%PQ%RS;%HLM%THU%%% 0H;%9;V9>%7W9H:;<FNXY<! y! };! I 1 = (x +1) 2 x 2 dx 2 3 ∫ = (1+ 2 x + 1 x 2 )dx 2 3 ∫ = (x + 2ln x − 1 x ) 3 2 = 7 6 + 2ln 3 2 *! };! I 2 = ln x x dx 2 3 ∫ = ln xd(lnx) 2 3 ∫ = 1 2 ln 2 x 3 2 = ln 2 3−ln 2 2 2 *! };! I = I 1 − I 2 = 7 6 + 2ln 3 2 − ln 2 3− ln 2 2 2 *!!!! 0tQ%B%uhn?%R;oJvX!#$%!$A3$![\3L!.T]!7U3L!^_#*^`_`#`!EM!7-H!^_#![&!.F'!L1-E!7aC!EJ3$!Fb! A 'B = a 7 2 *!Rc1!Bbdbe![f3![Og.![&!.TC3L!71='!E-E!EJ3$!^`_`b^`#`b##`*!@X3$!.$=!.XE$!h$)1![\3L! .T]!^_#*^`_`#`!5&!h$%,3L!E-E$!.i!71='!^!723!'j.!G$Q3L!:Bde;*!! ! };! S ABC = 1 2 AB.AC sin60 0 = a 2 3 4 *!!! @F'!L1-E!5CK3L!^`^_!EM•! ! AA'= A 'B 2 − AB 2 = 7a 2 4 − a 2 = a 3 2 *!! };! V ABC.A 'B 'C ' = AA '.S ABC = a 3 2 . a 2 3 4 = 3a 3 8 :75 ;*! };!Rc1!e`![&!.TC3L!71='!EIF!__`!.F!EM!ee`‡‡_`#`‡‡Bd!343! 'j.!G$Q3L!:Bde;![&!'j.!G$Q3L!:Bdee`;*! Rc1!rbr`bˆb‰![f3![Og.![&!.TC3L!71='!EIF!_`#`b_#bBdbrr`!.F! EM•!ˆ‰![&!L1F%!.CH23!EIF!'j.!G$Q3L!:^^`r`r;!5N1!'j.! G$Q3L!:Bde;*! •OP3L!.$Q3L!^r`!Ep.!ˆ‰!.J1!|b!Ep.!^`r!.J1!>*! @F!EM•!_#!5CK3L!LME!5N1!'j.!G$Q3L!:^^`r`r;!5&!ee`‡‡_#!343!ee`!5CK3L!LME!5N1!^>!:";*! Bj.!h$-E!.F'!L1-E!^^`r!5CK3L!EZ3!.J1!^!5A! AE = AA '= a 3 2 b!z%!7M!^>!5CK3L!LME!5N1!^`r!:<;*!! Bj.!h$-E!^`r‡‡ˆ‰!:‚;*! @i!:";b:<;b:‚;!(CH!TF•! AI ⊥ (MNP) ⇒ AH = d(A;(MNP)) *! };!q%!ˆ‰![&!7OP3L!.TC3L!0A3$!EIF!.F'!L1-E!^`r`r!343!|![&!.TC3L!71='!EIF!>r`b![J1!EM!>![&!.TC3L! 71='!EIF!^r`!343! AH = 3 4 AE ' = 3 4 AE 2 + EE ' 2 = 3 4 3a 2 4 + 3a 2 4 = 3a 6 8 *! v€H! d(A;(MNP))= 3a 6 8 *! 0ZMH%C>!!Rc1!r![&!.TC3L!71='!_#b!E$c3!$s!.T]E!.%J!7D!klHV!(F%!E$%•! A(0;0;0),E( a 3 2 ;0;0),B( a 3 2 ;− a 2 ;0),C( a 3 2 ; a 2 ;0),A'(0;0; a 3 2 ),B '( a 3 2 ;− a 2 ; a 3 2 ),C '( a 3 2 ; a 2 ; a 3 2 ) *! +$1!7M!.%J!7D!Bbdbe![f3![Og.![&! M( a 3 4 ;− a 4 ; a 3 2 ),N( a 3 4 ; a 4 ; a 3 2 ),P( a 3 2 ; a 2 ; a 3 4 ) *! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%THẦY%ĐẶNG%THÀNH%NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! "! #$%!&'(!!!! MN ! "!! = (0; a 2 ;0)/ /(0;1;0),MP ! "!! = ( a 3 4 ; 3a 4 ;− a 3 4 ) / /(1; 3;−1) )!*+!*,-!./!0120!.$%34!-5'! 6789:!;+! n (MNP) ! "!!!! = 1 0 3 −1 ; 0 0 −1 1 ; 0 1 1 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ = (−1;0;−1) <!! =>!*?%!01@/4A!.&>41!BC.!01D4A!6789:!;+! x + z− 3a 3 4 = 0 <! E'!-F(! d(A;(MNP))= − 3a 3 4 2 = 3a 6 8 <!!!! Câu%7%(1,0%điểm).!E&G4A!BC.!01D4A!*HI!.&J-!.GK!LM!NO%!-1G!L@P4A!.&Q4! (C ) : x 2 + y 2 = 1 <!RSI! T!;+!LIUB!.1$M-!L@P4A!.1D4A! y −3 = 0 !*+!V)W!;X4!;@Y.!;+!-2-!.I30!LIUB!-5'!.I30!.$%34!Z[!.\!T! L34!6W:<!]$'!V!Z[!L@P4A!.1D4A!^G4A!^G4A!*HI!TW!-_.!6W:!.KI!`)!T`!-_.!6W:!.KI!a<!E>B!.GK!LM! LIUB!T)!bI3.!Va!-_.!TW!.KI!LIUB!c6def:<! ! g@P4A!.&Q4!6W:!-F!.hB!;+!Ai-!.GK!LM!N6jej:)!b24!Zk41! bl4A!d<! Phát%hiện%tính%chất%hình%học:% E'!-F!c!;+!.&$4A!LIUB!TW! Chứng&minh.& cW!;+!.I30!.$%34!-5'!L@P4A!.&Q4!6W:!4m4! IC 2 = IE.IB (1) <! E1nG!.k41!-1o.!AF-!4pq!.I30!*+!AF-!.KG!brI!.I30!.$%34!*+! sh%!-$4A!.'!-F(! ! EBA ! = BDA ! BDA ! = EAI ! (so le) ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇒ EBA ! = EAI ! <! `G!LF!.'B!AI2-!cVT!Lt4A!sK4A!*HI!.'B!AI2-!cTa6A<A:<!! u:!#$%!&'(! IA IE = IB IA ⇒ IA 2 = IB.IE (2) <! E\!6d:!*+!6f:!^$%!&'(! IA = IC !)!1'%!c!;+!.&$4A!LIUB!TW<! u:!RSI!T6'ev:!.1$M-!L@P4A!.1D4A! y −3 = 0 )!sG!c!;+!.&$4A!LIUB!TW!4m4!W6fw'ed:<! 7C.!Z12-(! OC 2 = 1⇔ (2− a) 2 +1=1 ⇔ a = 2 ⇒ A(2;3) <! Kết%luận:!=?%!T6fev:!;+!LIUB!-X4!.>B<!!! Bình%luận:!E'!-F!.1U!-1x4A!BI41!c!;+!.&$4A!LIUB!TW!41@!^'$(! y[!.I30!.$%34!!*HI!6W:!.KI!`)!-_.!TW!.KI!7<!z>41!.1'4A!TV`7!-F! DBA ! = BDM ! !4m4!;+!1>41! .1'4A!-h4<!#$%!&'(! BAI ! = M " *+!W!;+!.&$4A!LIUB!-5'!T7<! RSI!y!;+!.&$4A!LIUB!-5'!T`)!.'!-F!Wy!;+!L@P4A!.&$4A!b>41!-5'!.'B!AI2-!T`7!4m4! ! CK = MD 2 = AB 2 = AC 2 <! E'B!AI2-!TVc!*+!WTy!-F! AB = AC ,ABI ! = CAK ! , BAI ! = ACK ! 4m4!bl4A!41'$<! 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! {! `G!LF! AI = CK )!*>!*?%! AI = AC 2 ⇒ I !;+!.&$4A!LIUB!-5'!TW<! Câu%8%(1,0%điểm).%E&G4A!Z1p4A!AI'4!*HI!1|!.&J-!.GK!LM!NO%}!-1G!1'I!LIUB!T6defewd:!*+!V6jejef:<! E>B!.GK!LM!LIUB!c!.1G~!B•4! IA !"! = 2IB !"! <!=I3.!01@/4A!.&>41!L@P4A!.1D4A! Δ LI!€$'!c)!-_.!.&J-!N}! *+!*$p4A!AF-!*HI!L@P4A!.1D4A!TV<!!! u:!=>! IA !"! = 2IB !"! 4m4!V!;+!.&$4A!LIUB!-5'!cT)!sG!LF!c6wdewfe":<! u:!RI~!^•!L@P4A!.1D4A! Δ -_.!.&J-!N}!.KI!LIUB!76jeje.:)!.'!-F(! IM ! "! = (1;2;t − 5) <! g@P4A!.1D4A!TV!-F!*,-!./!-1‚!01@/4A! AB ! "! = (−1;−2;3) <! `G!TV!*+!! Δ *$p4A!AF-!4m4(!! −1.1− 2.2+ 3(t − 5) = 0 ⇔ 3t −10= 0 ⇔ t = 10 3 ⇒ IM ! "! = (1;2;− 5 3 ) / /(3;6;−5) <! u:!g@P4A!.1D4A! Δ LI!€$'!c!*+!-F!*,-!./!-1‚!01@/4A!6ve{ew":!4m4!-F!01@/4A!.&>41!;+(! Δ : x +1 3 = y + 2 6 = z− 5 −5 <! Câu%9%(0,5%điểm).%E\!BM.!4Ah4!1+4A!fj!-h$!1ƒI)!.&G4A!LF!-F!„!-h$!1ƒI!Z1F!4A@PI!.'!Oh%!s…4A! .1+41!1'I!L†!.1I!B‡I!L†!.1I!AtB!dj!-h$)!*+!-2-!-h$!.&G4A!BM.!L†!L@Y-!L241!^i!.1x!.…!.\!Wh$!d! L34!Wh$!dj<!Ek41!O2-!^$o.!LU!Oh%!s…4A!L@Y-!1'I!L†!.1I!B+!B‡I!L†!.1I!L†$!AtB!f!-h$!1ƒI!Z1F<! y1p4A!AI'4!Bˆ$!;+!^i!-2-1!Oh%!s…4A!1'I!L†!.1I!B‡I!L†!.1I!AtB!dj!-h$!L@Y-!-1S4!&'!.\!4Ah4! 1+4A!fj!-h$!1ƒI<! u!W1S4!&'!dj!-h$!1ƒI!-1G!L†!.1x!41o.)!^'$!LF!^_0!O30!.1nG!.1x!.…!.\!-h$!d!L34!-h$!dj!-F! C 20 10 .10! !-2-1<! u!dj!-h$!-Q4!;KI!;o%!;+B!L†!.1x!1'I)!*+!^_0!O30!.1nG!.1x!.…!.\!-h$!d!L34!-h$!dj!-F!dj‰!-2-1<! =?%!Z1p4A!AI'4!Bˆ$! Ω = C 20 10 .10!.10!= (10!) 2 .C 20 10 <! u:!RSI!T!;+!bI34!-i!Oh%!s…4A!L@Y-!1'I!L†!.1I!B‡I!L†!AtB!f!-h$!1ƒI!Z1F<! u!W1S4!&'!f!-h$!1ƒI!Z1F!.&G4A!„!-h$)!*+!Š!-h$!1ƒI!s‹!.&G4A!d{!-h$!-1G!L†!.1x!41o.)!^'$!LF!^_0! O30!dj!-h$!4+%!.1nG!.1x!.…!.\!-h$!d!L34!-h$!dj!-F! C 4 2 .C 16 8 .10! -2-1<! u!dj!-h$!-Q4!;KI!;o%!;+B!L†!.1x!1'I)!*+!^_0!O30!.1nG!.1x!.…!.\!-h$!d!L34!-h$!dj!-F!dj‰!-2-1<! =?%!^i!01X4!.•!-5'!bI34!-i!T!;+! Ω A = C 4 2 .C 16 8 .10!.10!= (10!) 2 .C 4 2 .C 16 8 <! =?%!O2-!^$o.!-X4!.k41! P(A)= Ω A Ω = (10!) 2 .C 4 2 .C 16 8 (10!) 2 .C 20 10 = 135 323 <!!!! Câu%10%(1,0%điểm).!W1G!')b)-)B)4)0!;+!-2-!^i!.1…-!.1G~!B•4! a 2 + b 2 + c 2 = m 2 + n 2 + p 2 = 9 <!E>B! AI2!.&q!;H4!41o.!-5'!bIU$!.1x-! P = 9− a− 2b−2c + 9− m− 2n− 2p + 9− am− bn− cp <!! E'!-F(! P = (a−1) 2 + (b− 2) 2 + (c− 2) 2 2 + (m−1) 2 + (n− 2) 2 + (p− 2) 2 2 + (a− m) 2 + (b− n) 2 + (c− p) 2 2 <! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%THẦY%ĐẶNG%THÀNH%NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! Œ! u:!E&G4A!Z1p4A!AI'4!NO%})!O,.!b'!LIUB!T6'ebe-:)!V6Be4e0:)!W6defef:)!.'!-F!T)V)W!-•4A!.1$M-! BC.!-X$!6#:!-F!01@/4A!.&>41( x 2 + y 2 + z 2 = 9 )!b24!Zk41! R = 3 <!! =+! P = 1 2 (AB + BC +CA) )!.'!ASI!69:!;+!BC.!01D4A!-1x'!.'B!AI2-!TVW<!y1I!LF!69:!-_.!6#:!.1nG! AI'G!.$%34!B+!BM.!L@P4A!.&Q4!b24!Zk41! R '= R 2 − d 2 (O;(P)) ≤ R <!!! u:!E'!-F(! AB + BC +CA ≤ 3 3R' ≤ 3 3R = 9 3 ⇒ P ≤ 9 6 2 <! `o$!bl4A!O~%!&'!Z1I!*+!-1‚!Z1I!.'B!AI2-!TVW!L†$!*+!N!;+!.&S4A!.hB!-5'!.'B!AI2-!TVW<!! `G!;$p4!-F!BC.!01D4A!69:!LI!€$'!N)W!*+!69:!-_.!6#:!.1nG!AI'G!.$%34!;+!BM.!L@P4A!.&Q4!6W:)! .&m4!6W:!;$p4!.>B!L@Y-!1'I!LIUB!T)V!LU!.'B!AI2-!TVW!4m4!so$!bl4A!O~%!&'<!! =?%!AI2!.&q!;H4!41o.!-5'!9!bl4A! 9 6 2 <!!!! Bình%luận:%WF!.1U!-J!.1U!1G2!so$!bl4A!41@!^'$(! a 2 + b 2 + c 2 = 9 (a−1) 2 + (b− 2) 2 + (c− 2) 2 = 27 m 2 + n 2 + p 2 = 9 (m−1) 2 + (n− 2) 2 + (p− 2) 2 = 27 a + m+1 3 = b + n+ 2 3 = c+ p+ 2 3 = 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ <! u:!ŽPI!AI~I!.&m4!L•!^•!b+I!.G24!-…-!.&q!1>41!1S-!^'$(! Tam giác ABC nội tiếp đường tròn (O) cho trước có chu vi lớn nhất khi ABC là tam giác đều. Chứng minh. Gọi R là bán kính đường tròn, theo định lý hàm số Sin ta có: BC = 2Rsin A,CA = 2RsinB,AB = 2RsinC . Vì vậy chu vi tam giác ABC là P = 2R(sin A + sinB + sinC) (1) , Mặt khác: sin A + sinB + sinC = 2sin A + B 2 cos A − B 2 + 2sin C 2 cos C 2 ≤ 2cos C 2 1+ sin C 2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ (2) . #•!sJ4A!bo.!LD4A!.1x-!T7!•R7!-1G!b'!^i!.1…-!Z1p4A!hB!.'!-F(! 2cos C 2 1+ sin C 2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ = 2. 2cos 2 C 2 (1+ sin C 2 )(1+ sin C 2 ) ≤ 2. 2cos 2 C 2 +1+ sin C 2 +1+ sin C 2 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ 3 = 2. 9 2 −2(sin C 2 − 1 2 ) 2 3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ 3 ≤ 2. ( 3 2 ) 3 = 3 3 2 (3) <!! Từ (1),(2),(3) ta có: P ≤ 3 3R . Dấu bằng xảy ra khi và chỉ khi A = B sin C 2 = 1 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇔ A = B = C = 60 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! Š! Cách&2:!#•!sJ4A!bo.!LD4A!.1x-!W'$-1%!•#-1•'&}!.'!-F!! P = 1 2 (AB + BC +CA)≤ 3 2 (AB 2 + BC 2 + CA 2 ) = 3 2 (OB ! "! −OA ! "! ) 2 + (OC ! "!! −OB ! "! ) 2 + (OA ! "! −OC ! "!! ) 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 3 2 (6R 2 −2(OA ! "! .OB ! "! + OC ! "!! .OB ! "! + OA ! "! .OC ! "!! ) <! u:!7C.!Z12-!ASI!R!;+!.&S4A!.hB!.'B!AI2-!TVW!.'!-F! ! OA ! "! + OB ! "! + OC ! "!! = 3OG ! "!! ⇒ (OA ! "! + OB ! "! + OC ! "!! ) 2 = 9OG 2 ⇔ 3R 2 + 2(OA ! "! .OB ! "! + OC ! "!! .OB ! "! + OA ! "! .OC ! "!! )= 9OG 2 ⇒ 2(OA ! "! .OB ! "! + OC ! "!! .OB ! "! + OA ! "! .OC ! "!! )= 9OG 2 − 3R 2 <! =>!*?%! P ≤ 3 2 (6R 2 −(9OG 2 −3R 2 )) = 27R 2 2 − 27OG 2 2 ≤ 27R 2 2 = 9 6 2 <! `o$!bl4A!O~%!&'!Z1I!*+!-1‚!Z1I! O ≡ G AB = BC = CA ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ <!!!! Bài%tập%tương%tự%w% Bài số 01. Cho a,b,c,d là các số thực thoả mãn a 2 + b 2 = c 2 + d 2 = 10 . Tìm giá trị lớn nhất của biểu thức P = 10− a − 3b + 10−c −3d + 10− ac−bd . Đ/s: P max = 3 15 . Bài số 02. Cho a,b,c,d là các số thực thoả mãn a 2 + b 2 = c 2 + d 2 = 25 . Tìm giá trị lớn nhất của biểu thức P = 25− 3a − 4b + 25−3c− 4d + 25− ac−bd . Đ/s: P max = 15 6 2 . ! . 3 4 ; a 4 ; a 3 2 ),P( a 3 2 ; a 2 ; a 3 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! "! #$%!&'(!!!! . BAI ! = ACK ! 4m4!bl4A!41'$<! 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! {! `G!LF! AI. 2) 2 2 + (a− m) 2 + (b− n) 2 + (c− p) 2 2 <! 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! Œ! u:!E&G4A!Z1p4A!AI'4!NO%})!O,.!b'!LIUB!T6'ebe-:)!V6Be4e0:)!W6defef:)!.'!-F!T)V)W!-•4A!.1$M-! BC.!-X$!6#:!-F!01@/4A!.&>41(