KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%:%THẦY:%ĐẶNG%THÀNH%NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! "! Khoá%giải%đề%THPT%Quốc%Gia%–%Thầy:%Đặng%Thành%Nam% Môn:%Toán;%ĐỀ%SỐ%18/50% Ngày%thi%:%22/03/2015% Thời%gian%làm%bài:%180%phút,%không%kể%thời%gian%giao%đề% Liên%hệ%đăng%ký%khoá%học%–%Hotline:%0976%266%202%–%Chi%tiết:%www.mathlinks.vn%% Câu%1%(2,0%điểm).!#$%!$&'!()! y = x 3 − x 2 + 4 (1) *!! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*! <* =>'!.%?!7@!71A'!B!.$C@D!:";!(E%!D$%!.12F!.CG23!DHE!:";!.?1!B!.?%!5I1!$E1!.JKD!.%?!7@!'@.! .E'!L1-D!DM3*! Câu%2%(1,0%điểm).% E; N1,1!F$OP3L!.J>3$! 2cos 2 x + 3 sin 2x = 2sin 5x +1 *! 0; N1,1!0Q.!F$OP3L!.J>3$! log x.log(100x 2 ) = 4 *!!!! Câu%3%(1,0%điểm).!=R3$!.RD$!F$M3! I = 1− x 2 ln x x dx 1 2 ∫ *! Câu%4%(1,0%điểm).% E; S-D!793$!F$T3!,%!DHE!()!F$UD! z = (3+ 2i )(−3+ i)+ 1−3i 2− i *!! 0; #$%!V$E1!.J1A3! (1+ 2x ) n = a 0 + a 1 x + a 2 x 2 + + a n x n *!=>'!$W!()!DHE!()!$?3L!D$UE! x 8 !.J%3L! V$E1!.J1A3!012.! a 0 + 9a 1 = 2a 2 +1 *! Câu%5%(1,0%điểm).!#$%!$>3$!D$XF!Y*Z[#\!DX!7-G!Z[#\!]&!$>3$!.$%1^ AB = a 3,BD = 3a *!_>3$! D$12C!5C`3L!LXD!DHE!Y!.J43!'a.!F$b3L!:Z[#\;!.Jc3L!5I1!.JC3L!71A'!D?3$!Z#*![12.!D`(13!LXD! .?%!0d1!'a.!F$b3L!:YZ[;!5&!'a.!F$b3L!:Z[#\;!0e3L! 21 7 *!=R3$!.$f%!E!.$A!.RD$!V$)1!D$XF! Y*Z[#\!5&!0-3!VR3$!'a.!DTC!3L%?1!.12F!.U!g1W3!Y*Z#\*!! Câu%6%(1,0%điểm).!=J%3L!V$`3L!L1E3!5I1!$W!.JKD!.%?!7@!hiGj!D$%!71A'!Z:"klkm";!5&!'a.!F$b3L! (P ) : 2x − 2y + z + 6 = 0 *!n12.!F$OP3L!.J>3$!7Oo3L!.$b3L!g!71!pCE!Z!5&!5C`3L!LXD!5I1!'a.! F$b3L!:q;*!=>'!.%?!7@!71A'!B!.$%,!'r3! AM = 2 5&!V$%,3L!D-D$!.s!B!723!:q;!7?.!L1-!.J9!]I3! 3$Q.*! Câu%7%(1,0%điểm).!=J%3L!'a.!F$b3L!5I1!.JKD!.%?!7@!hiG!D$%!.E'!L1-D!L1-D!Z[#!3@1!.12F!7Oo3L! .Jt3! (C ) : x 2 + (y− 29 8 ) 2 = 697 64 *!=>'!.%?!7@!7u3$!Z^!012.!7Oo3L!.Jt3!3L%?1!.12F!.E'!L1-D!v[#!Dw.! Z[^Z#!]T3!]Ox.!.?1!D-D!71A'! M ( 21 10 ;1), N (0; 13 5 ),(M ≠ B, N ≠ C ) !5&!Z!DX!$%&3$!7@!M'*! Câu%8%(1,0%điểm).!N1,1!$W!F$OP3L!.J>3$! (x 2 − xy +1)( y 2 − xy +1) = 1 1 x 2 + 3 + x 2 − x +1 = y + 1 y +1 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ (x, y ∈ !) *! Câu%9%(1,0%điểm).%#$%!i^G^j!]&!D-D!()!F$UD!.$%,!'r3! x = y = z = 1 *!=>'!L1-!.J9!]I3!3$Q.!DHE! 01AC!.$UD! P = x + y − z + y + z − x + z + x − y *! xxxHẾTxxx KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%:%THẦY:%ĐẶNG%THÀNH%NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! <! PHÂN%TÍCH%BÌNH%LUẬN%ĐÁP%ÁN% Câu%1%(2,0%điểm).!#$%!$&'!()! y = x 3 − x 2 + 4 (1) *!! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*! <* =>'!.%?!7@!71A'!B!.$C@D!:";!(E%!D$%!.12F!.CG23!DHE!:";!.?1!B!.?%!5I1!$E1!.JKD!.%?!7@!'@.! .E'!L1-D!DM3*! "* _yD!(13$!./!L1,1*! <* Ny1! M (m;m 3 − m 2 + 4) ]&!71A'!DT3!.>'*! z;!=12F!.CG23!.?%!5I1!$E1!.JKD!.%?!7@!'@.!.E'!L1-D!DM3!343!.12F!.CG23!5C`3L!LXD!5I1!7Oo3L! F$M3!L1-D!DHE!$E1!.JKD!.%?!7@k!! z;!q$OP3L!.J>3$!$E1!7Oo3L!F$M3!L1-D!DHE!$E1!.JKD!.%?!7@!]&! y = −x $%aD! y = x *! \%!7X!$W!()!LXD!DHE!.12F!.CG23!.?1!B!]&! k = y'(m) = 3m 2 − 2m = ±1 ⇔ 3m 2 − 2m −1= 0 3m 2 − 2m +1= 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⇔ m =1 m = − 1 3 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ *! Kết%luận:!#X!$E1!71A'!.$%,!'r3!]&! M (1;3) 5&! M (− 1 3 ; 104 27 ) *!!!!!!! Câu%2%(1,0%điểm).% E; N1,1!F$OP3L!.J>3$! 2cos 2 x + 3 sin 2x = 2sin 5x +1 *! 0; N1,1!0Q.!F$OP3L!.J>3$! log x.log(100x 2 ) = 4 *!!!! E; q$OP3L!.J>3$!.OP3L!7OP3L!5I1{! ! cos2x +1+ 3 sin2x = 2sin5x +1 ⇔ 3 sin 2x + cos2x = 2sin5x ⇔ 3 2 sin2x + 1 2 cos2x = sin5x ⇔ sin(2x + π 6 ) = sin5x ⇔ 5x = 2x + π 6 + k2π 5x = 5π 6 − 2x + k2π ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⇔ x = π 18 + k 2π 3 x = 5π 42 + k 2π 7 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ *! Kết%luận:!q$OP3L!.J>3$!DX!3L$1W'!]&! x = π 18 + k 2π 3 ;x = 5π 42 + k 2π 7 ,k ∈ ! *!!! 0; |1}C!V1W3{! x > 0 *! q$OP3L!.J>3$!.OP3L!7OP3L!5I1{! ! log x.(log100+ log x 2 ) = 4 ⇔ log x.(2+ 2log x ) = 4 ⇔ log 2 x + log x −2 = 0 ⇔ log x = 1 log x = −2 ⎡ ⎣ ⎢ ⎢ ⇔ x = 10 x = 1 100 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ *! Kết%luận:!q$OP3L!.J>3$!DX!3L$1W'! x = 10;x = 1 100 *!!!! Câu%3%(1,0%điểm).!=R3$!.RD$!F$M3! I = 1− x 2 ln x x dx 1 2 ∫ *! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%:%THẦY:%ĐẶNG%THÀNH%NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! ~! =E!DX{!! I = ( 1 x − x ln x )dx 1 2 ∫ = ln x 2 1 − x ln x dx 1 2 ∫ = ln 2−K *! z;!|a.! u = ln x dv = xdx ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ du = dx x v = x 2 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⇒ K = x 2 2 ln x 2 1 − xdx 2 1 2 ∫ = 2ln 2− x 2 4 2 1 = 2ln 2− 3 4 *! Kết%luận:!n•G! I = ln2− (2ln 2− 3 4 ) = 3 4 − ln2 *!!! Câu%4%(1,0%điểm).% E; S-D!793$!F$T3!,%!DHE!()!F$UD! z = (3+ 2i )(−3+ i)+ 1−3i 2− i *!! 0; #$%!V$E1!.J1A3! (1+ 2x ) n = a 0 + a 1 x + a 2 x 2 + + a n x n *!=>'!$W!()!DHE!()!$?3L!D$UE! x 8 !.J%3L! V$E1!.J1A3!012.! a 0 + 9a 1 = 2a 2 +1 *! E; =E!DX{! z = (−11−3i) + (1− 3i )(2+ i ) 5 = −11−3i + 5− 5i 5 = −11−3i +1− i = −10− 4i *! n•G!F$T3!,%!DHE!j!0e3L!m€*! 0; =E!DX{! (1+ 2x ) n = C n k .2 k .x k k=0 n ∑ = a 0 + a 1 x + a 2 x 2 + + a n x n ⇒ a 0 = C n 0 .2 0 = 1 a 1 = C n 1 .2 1 = 2n a 2 = C n 2 .2 2 = 2n(n −1) ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ *! =$f%!L1,!.$12.!.E!DX{! ! 1+ 9.2n = 2n(n −1) +1 ⇔ 2n(n −10) = 0 ⇔ n = 10 *! YCG!JE!$W!()!DHE!()!$?3L!D$UE! x 8 !]&! a 8 = C 10 8 .2 8 *!!! Câu%5%(1,0%điểm).!#$%!$>3$!D$XF!Y*Z[#\!DX!7-G!Z[#\!]&!$>3$!.$%1^ AB = a 3,BD = 3a *!_>3$! D$12C!5C`3L!LXD!DHE!Y!.J43!'a.!F$b3L!:Z[#\;!.Jc3L!5I1!.JC3L!71A'!D?3$!Z#*![12.!D`(13!LXD! .?%!0d1!'a.!F$b3L!:YZ[;!5&!'a.!F$b3L!:Z[#\;!0e3L! 21 7 *!=R3$!.$f%!E!.$A!.RD$!V$)1!D$XF! Y*Z[#\!5&!0-3!VR3$!'a.!DTC!3L%?1!.12F!.U!g1W3!Y*Z#\*!! Ny1!_!]&!L1E%!71A'!DHE!Z#!5&![\^!.E!DX! SH ⊥ (ABCD ) *! •F!gK3L!793$!]‚!$&'!()!D`(13!D$%!.E'!L1-D!Z[\!.E!DX{! ! cos BAD ! = AB 2 + AD 2 − BD 2 2AB.AD = 3a 2 + 3a 2 − 9a 2 2.a 3.a 3 = − 1 2 ⇒ BAD ! = 120 0 *! \%!7X!D-D!.E'!L1-D!Z[#^!Z\#!7}C!D?3$! a 3 *! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%:%THẦY:%ĐẶNG%THÀNH%NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! €! ! +ƒ!_„!5C`3L!LXD!5I1!Z[!.?1!„^!.E!DX{! AB ⊥ HN AB ⊥ SH ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ AB ⊥ (SHN ) ⇒ ((SAB );(ABCD )) ! = SNH ! *! =E'!L1-D!5C`3L!Z[_!DX{! ! HN = AH .sin60 0 = a 3 2 . 3 2 = 3a 4 *! =E'!L1-D!5C`3L!Y_„!DX{! cosSNH ! = 21 7 ⇒ tanSNH ! = 2 3 ⇒ SH = HN .tanSNH ! = a 3 2 *!!! S ABCD = 4S HAB = 2HN .AB = 2.a 3. 3a 4 = 3a 2 3 2 *! YCG!JE{! V S . ABCD = 1 3 SH .S ABCD = 1 3 . a 3 2 . 3a 2 3 2 = 3a 3 4 :75 ;*! z;!=E!DX{! HS = HA = HC = a 3 2 !343!_!]&!.M'!7Oo3L!.Jt3!3L%?1!.12F!.E'!L1-D!YZ#*! Ba.!V$-D![\!5C`3L!LXD!5I1!'a.!F$b3L!:YZ#;!343![\!]&!.JKD!7Oo3L!.Jt3!3L%?1!.12F!.E'!L1-D! YZ#*! Ny1!h!]&!.M'!.E'!L1-D!7}C!Z#\!.E!DX! OA = OC = OD = OS = a 343!h!]&!.M'!'a.!DTC!3L%?1!.12F! .U!g1W3!Y*Z#\!5&!0-3!VR3$!0e3L!E*!!!! Câu%6%(1,0%điểm).!=J%3L!V$`3L!L1E3!5I1!$W!.JKD!.%?!7@!hiGj!D$%!71A'!Z:"klkm";!5&!'a.!F$b3L! (P ) : 2x − 2y + z + 6 = 0 *!n12.!F$OP3L!.J>3$!7Oo3L!.$b3L!g!71!pCE!Z!5&!5C`3L!LXD!5I1!'a.! F$b3L!:q;*!=>'!.%?!7@!71A'!B!.$%,!'r3! AM = 2 5&!V$%,3L!D-D$!.s!B!723!:q;!7?.!L1-!.J9!]I3! 3$Q.*! z;!|Oo3L!.$b3L!g!5C`3L!LXD!5I1!:q;!343!3$•3!5.F.!DHE!:q;!]&'!5…D!.P!D$u!F$OP3L^!(CG! JE{ u ! d = (2;−2;1) *!\%!7X{! d : x −1 2 = y −2 = z +1 1 *! z;!|A!V$%,3L!D-D$!.s!B!723!:q;!]I3!3$Q.!V$1!B!3e'!.J43!7Oo3L!.$b3L!g!5&!B!DX!V$%,3L!D-D$! 723!:q;!]I3!$P3!V$%,3L!D-D$!.s!Z!723!:q;^!V$1!7X!! M (1+ 2t;−2t;−1+ t ) *! =E!DX{! AM = 2 ⇔ 4t 2 + 4t 2 + t 2 = 4 ⇔ t = − 2 3 t = 2 3 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⇒ M ( 7 3 ;− 4 3 ;− 1 3 ) M (− 1 3 ; 4 3 ;− 5 3 ) ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ *! Kiểm%tra:% z;!nI1! M ( 7 3 ;− 4 3 ;− 1 3 ) ⇒ d(M ;(P )) = 13 3 *! z;!nI1! M (− 1 3 ; 4 3 ;− 5 3 ) ⇒ d(M ;(P )) = 1 3 *! Y%!(-3$!$E1!.JOo3L!$xF!.E!DX!71A'!DT3!.>'!]&! M ( 7 3 ;− 4 3 ;− 1 3 ) *! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%:%THẦY:%ĐẶNG%THÀNH%NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! †! Kết%luận:!n•G! M ( 7 3 ;− 4 3 ;− 1 3 ) *!!!!!!! Cách%2:!Ny1! M (a;b;c ) ⇒ AM 2 = 4 ⇔ (a −1) 2 + b 2 + (c +1) 2 = 4 *! =E!DX{! ! d (M ;(P )) = 2a −2b + c + 6 3 = 2(a −1)− 2b + (c +1)+ 7 3 ≤ 2(a −1)− 2b + (c +1) + 7 3 ≤ (2 2 + (−2) 2 +1 2 )((a − 1) 2 + b 2 + (c +1) 2 ) + 7 3 = 13 3 *! \QC!0e3L!i,G!JE!V$1!5&!D$u!V$1! (a −1) 2 + b 2 + (c +1) 2 = 4 a −1 2 = b −2 = c +1 1 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇔ a = 7 3 ,b = − 4 3 ,c = − 1 3 *! =E!DX!V2.!pC,!.OP3L!./*!! Câu%7%(1,0%điểm).!=J%3L!'a.!F$b3L!5I1!.JKD!.%?!7@!hiG!D$%!.E'!L1-D!L1-D!Z[#!3@1!.12F!7Oo3L! .Jt3! (C ) : x 2 + (y− 29 8 ) 2 = 697 64 *!=>'!.%?!7@!7u3$!Z^!012.!7Oo3L!.Jt3!3L%?1!.12F!.E'!L1-D!v[#!Dw.! Z[^Z#!]T3!]Ox.!.?1!D-D!71A'! M ( 21 10 ;1), N (0; 13 5 ),(M ≠ B, N ≠ C ) !5&!Z!DX!$%&3$!7@!M'*! ! +)%Phát%hiện%tính%chất%hình%học:%% =E!DX!Zv!5C`3L!LXD!5I1!B„^!.$•.!5•G{! Ny1!_!]&!L1E%!71A'!DHE!Zv!5&!B„^!.E!DX{! =$f%!.R3$!D$Q.!LXD!d!.M'!.E!DX{! ! AIC ! = 2ABC ! ⇒ IAC ! = 90 0 − ABC ! *! z;!=U!L1-D![B„#!3@1!.12F!7Oo3L!.Jt3!343! ANH ! = ABC ! *! #@3L!]?1!.$f%!52!.E!7OxD{! ! IAC ! + ACH ! = 90 0 ⇒ AHC ! = 90 0 *! n•G!Zv!5C`3L!LXD!5I1!B„*!! z;!! ! ! z;!|Oo3L!.$b3L!Zv!71!pCE!v!5&!5C`3L!LXD!5I1!B„!DX!F$OP3L!.J>3$!]&! 21x −16y + 58 = 0 *! =%?!7@!71A'!Z!]&!3L$1W'!DHE!$W! 21x −16y + 58 = 0 x 2 + (y− 29 8 ) 2 = 697 64 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇔ x = −2, y =1 x = 2, y = 25 4 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⇒ A(−2;1);A(2; 25 4 ) *! Kết%luận:!n•G!Z:m<k";!$%aD!Z:<k<†‡€;*!!! Câu%8%(1,0%điểm).!N1,1!$W!F$OP3L!.J>3$! (x 2 − xy +1)( y 2 − xy +1) = 1 1 x 2 + 3 + x 2 − x +1 = y + 1 y +1 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ (x, y ∈ !) *! q$OP3L!.J>3$!.$U!3$Q.!DHE!$W!.OP3L!7OP3L!5I1{! ! (x 2 − xy +1)( y 2 − xy +1) = 1 ⇔ (x − y) 2 (xy −1) = 0 ⇔ x = y xy =1 ⎡ ⎣ ⎢ ⎢ *! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%:%THẦY:%ĐẶNG%THÀNH%NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! ˆ! #,!$E1!.JOo3L!$xF^!=$EG!5&%!F$OP3L!.J>3$!.$U!$E1!DHE!$W!.E!7OxD{! 3+ 1 x 2 + x 2 − x +1 = x + 1 x +1 *! [>3$!F$OP3L!$E1!52!DHE!F$OP3L!.J>3$!.E!7OxD{! ! 3+ 1 x 2 + x 2 − x +1+ 2 (3+ 1 x 2 )(x 2 − x +1) = x 2 + 1 x 2 + 3+ 2 x + 2x ⇔ 2 (3+ 1 x 2 )(x 2 − x +1) = 3x + 2 x −1 !*! D$‰!‚!(Š!gK3L!0Q.!7b3L!.$UD!#ECD$G!‹YD$ŒEJ.j!.E!DX{! ! 2 (3+ 1 x 2 )(x 2 − x +1) = (3+ 1 x 2 )((2− x ) 2 + 3x 2 ) ≥ 2− x x + 3x ⇒VT (*) ≥VP (*) *! n>!5•G!gQC!0e3L!F$,1!i,G!JE! ⇔ 2− x 1 x = 3x 3 > 0 ⇔ x = 1 *! Kết%luận:!_W!q$OP3L!.J>3$!DX!3L$1W'!gCG!3$Q.! (x; y) = (1;1) *!! Bình%luận:!=E!DX!.$A!0>3$!F$OP3L!$E1!52!7A!L1,1!F$OP3L!.J>3${! ! 2 (3+ 1 x 2 )(x 2 − x +1) = 3x + 2 x −1 *! =$•G!5•G!0>3$!F$OP3L!$E1!52!.E!7OxD{ (x −1)(5x 3 + 3x 2 + 8) = 0 (**) *!! #$‰!‚!„2C!i•l!.$>! x 2 − x +1 >1, x + 1 x +1<1 F$OP3L!.J>3$!5`!3L$1W'*! n>!5•G!! (**) ⇔ x = 1 *!!! Bài%tập%tương%tự%x!N1,1!$W!F$OP3L!.J>3$! x 2 − xy +1 + y 2 − xy +1 = (x − y) 2 + 4 1 x 2 + 3 + x 2 − x +1 = y + 1 y +1 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ *! Câu%9%(1,0%điểm).%#$%!i^G^j!]&!D-D!()!F$UD!.$%,!'r3! x = y = z = 1 *!=>'!L1-!.J9!]I3!3$Q.!DHE! 01AC!.$UD! P = x + y − z + y + z − x + z + x − y *! =$f%!L1,!.$12.!.E!DX{ x z = y z = 1 *!! +$1!7X! P = z .( x z + y z −1 + y z +1− x z + 1+ x z − y z ) = x z + y z −1 + y z +1− x z + 1+ x z − y z *! =E!7a.! a = x z ,b = y z ⇒ P = a + b −1 + b +1− a + 1+ a− b , a = b =1 *! YCG!JE!.83!.?1!D-D!LXD! 0 ≤ α ≤ β < 2π (E%!D$%! a = cosα + i.sinα,b = cosβ +i.sin β *! +$1!7X{! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%:%THẦY:%ĐẶNG%THÀNH%NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! Ž! ! a + b −1 = (cosα + cosβ −1) 2 + (sinα + sin β) 2 = 3− 2cosα −2cosβ + 2cosα cosβ + 2sinαsin β = 3− 4cos α + β 2 .cos α− β 2 + 2cos(α− β) = 1− 4cos α + β 2 .cos α− β 2 + 4cos 2 α− β 2 = 1− 4cosu.cosv + 4cos 2 v *! nI1! u = α + β 2 ,v = α− β 2 *! =OP3L!./!.E!DX{! ! b +1− a = 1− 4sinu.sinv + 4sin 2 v , a +1−b = 1+ 4sinu.sinv + 4sin 2 v *! \%!7X!(Š!gK3L!0Q.!7b3L!.$UD!#ECD$G!‹YD$ŒEJj!.E!DX{! ! ! P = 1− 4cosu.cosv + 4cos 2 v + 1− 4sinu.sinv + 4sin 2 v + 1+ 4sinu.sinv + 4sin 2 v ≤ 3(7− 4 cosu.cosv + 4sin 2 v) ≤ 21+12cosv +12sin 2 v = 33+12cosv −12cos 2 v = 36−12(cosv − 1 2 ) 2 ≤ 6 *! \QC!0e3L!i,G!JE!V$1!5&!D$u!V$1! cosv = 1 2 cosu = −1 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇔ α + β = 2π β −α = 2π 3 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇔ α = 2π 3 β = 4π 3 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ *! n•G!L1-!.J9!]I3!3$Q.!DHE!q!0e3L!ˆ*!! Bình%luận:!•!.Od3L!0&1!.%-3!.JOID!.143!]&!L1,'!()!0123!F$UD!5&!(EC!7X!D$CGA3!.s!g?3L!()!F$UD! (E3L!g?3L!7?1!()!7A!-F!gK3L!D-D!0Q.!7b3L!.$UD!ZB!‹NB!$%aD!#ECD$G!‹YD$ŒEJ.j!! z;!q$…F!D$CGA3!.s!()!F$UD!(E3L!g?3L!7?1!()!g/E!5&%!g?3L!]Ox3L!L1-D!DHE!()!F$UD*! z;!=E!DT3!D$‰!‚!$E1!D`3L!.$UD!]143!pCE3!723!'`7C3!DHE!()!F$UD!3$O!(EC{! ! z 1 z 2 = z 1 z 2 , z 1 .z 2 = z 1 . z 2 *!! z;![&1!.%-3!3&G!7a.!G4C!DTC!5}!N=•„!343!.Q.!3$143!3L$‘!723!51WD!(Š!gK3L!0Q.!7b3L!.$UD!#ED$G! ‹YD$ŒEJ.j!g?3L{! ! x 1 + x 2 + + x n ≤ n(x 1 + x 2 + + x n ) *!! z;!„2C!D-D!0&1!.%-3!G4C!DTC!5}!L1-!.J9!3$’!3$Q.!]‰D!7X!D$‰!‚!(Š!gK3L!0Q.!7b3L!.$UD!B13D%F(V1! g?3L{! ! x 1 2 + y 1 2 + x 2 2 + y 2 2 + + x n 2 + y n 2 ≥ (x 1 + x 2 + + x n ) 2 + (y 1 + y 2 + + y n ) 2 !*! \QC!0e3L!7?.!.?1! x 1 y 1 = x 2 y 2 = = x n y n *!! Bài%tập%tương%tựx%#$%!i^G^j!]&!D-D!()!F$UD!.$%,!'r3! x = y = z = 1 *!=>'!L1-!.J9!]I3!3$Q.!DHE! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%:%THẦY:%ĐẶNG%THÀNH%NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! “! 01AC!.$UD! P = x + y − z + y + z − x + z + x − y *! . KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%:%THẦY:%ĐẶNG%THÀNH %NAM% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! "! Khoá %giải %đề% THPT%Quốc%Gia%– %Thầy: %Đặng %Thành% Nam% Môn: %Toán; %ĐỀ%SỐ%18/50% Ngày %thi% :%22/03/2015% Thời%gian%làm%bài:%180%phút,%không%kể%thời%gian%giao %đề% Liên%hệ%đăng%ký%khoá%học%–%Hotline:%0976%266%202%–%Chi%tiết:%www.mathlinks.vn%% Câu%1%(2,0%điểm).!#$%!$&'!()! . Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! "! Khoá %giải %đề% THPT%Quốc%Gia%– %Thầy: %Đặng %Thành% Nam% Môn: %Toán; %ĐỀ%SỐ%18/50% Ngày %thi% :%22/03/2015% Thời%gian%làm%bài:%180%phút,%không%kể%thời%gian%giao %đề% Liên%hệ%đăng%ký%khoá%học%–%Hotline:%0976%266%202%–%Chi%tiết:%www.mathlinks.vn%% Câu%1%(2,0%điểm).!#$%!$&'!()! . 1 *!=>'!L1-!.J9!]I3!3$Q.!DHE! 01AC!.$UD! 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