!"#$%&%%&!'()*+,$ % /0)1!2$ !32*24)5+ 6+2$%7!81Không kể thời gian phát đề 9:; (Đề thi có 01 trang) 9 +2$2 điểm !"#$%&'()*+,- !./0 ! 1"2,3'4 , ! ,54'61 +2%$ điểm 7484%9#2,:%9#2'6/ !; 0<0 = ! − = + = x 2y 4 2x 3y 1 +2<$ điểm >?&@436A'6/ ! ( ) = +A 32 3 18 : 2 ! − + = − − + 15 12 6 2 6 B 5 2 3 2 +2=$< điểm #BC)@DE>,43F,54)F>1"GF,-'4 HI%6IFJ*FI#B),54J*@HI%431 !"EF)J !"GF,-@6IFKLI#B)@6IFKLDM4 N6'C)17&4.#643O'PKLQJRKLP4S1 '!A4<43)Q.QJQFT6((#B1 !A4FK1FL;> 1 ! = R OH 2 *U(4P.SV>1 WWWWWWWWWWWWWWWWWW.X"WWWWWWWWWWWWWWWWW >?@9 A/B$%&%%CDE$ % /F$ +2$2 điểm ! =YQY= ZO'4N6'=Q,Q= !ZO',0 ! '4 ,54'6D4,[D4/ '' ! , ≠ ! ⇔ − = − ≠ ⇔ = − ≠ − 1 1 2m vaø 2 n m vaø n 2 2 +2%$ điểm 7484%9#2,:%9#2'6/ !; 0<0 =';Q<Q ∆ = − = − = > ⇒ ∆ = = 2 2 b 4ac 4 4.3.1 4 0 4 2 K9#24:%C4:/ − − ∆ − − − + ∆ − + = = = − = = = − 1 2 b 4 2 b 4 2 1 x 1;x 2a 2.3 2a 2.3 3 @D@/'0;<0 = K9#24:%C4:/ = − = − = − 1 2 c 1 x 1;x a 3 ! − = − = − = = − ⇔ ⇔ ⇔ + = + = + = = x 2y 4 2x 4y 8 7y 7 y 1 2x 3y 1 2x 3y 1 2x 3y 1 x 2 \:%9#2(4:1"%4: ( ) { } = − S 2; 1 +2<$ điểm >?&@436A'6/ ! ( ) ( ) = + = + = =A 32 3 18 : 2 4 2 9 2 : 2 13 2 : 2 13 ! ( ) ( ) − + − + = − = − = − − + − + 3 5 4 12 3 2 15 12 6 2 6 B 3 5 2 3 2 5 2 3 2 +2=$< điểm !FJHI%6IO') · ⇒ = o ABO 90 ABO∆ ,6MP4J)F)J ] ABO∆ ^''4@_6P)F · ⇒ = o AOB 60 @D@/ ABO ∆ ,6MP4J · · = = = ⇒ = o OB R 1 cosAOB AOB 60 OA 2R 2 !'!.#643O'KL ⇒ ⊥OH PQ P4. "A4@).FJ · · + = + = o o o ABO AHO 90 90 180 ]A4@).FJ(4HI%1 \<43)Q.QJQFT6((#B1 !`a ∆ABP , ∆AQB µ A 6 · · » = = 1 ABP AQB sñBP 2 Pb4H'HI%6I,C6,54 (4HI%TR(6 ] ∆ ∆ −ABP AQB(g g)∽ ⇒ = ⇒ = 2 AB AP AP.AQ AB (1) AQ AB c$D@ ABO ∆ ,6MP4J*VEK4W'W "' ( ) = + ⇒ = − = − = 2 2 2 2 2 2 2 2 2 OA AB OB AB OA OB 2R R 3R (2) "G , ⇒ = 2 AP.AQ 3R ! AHO ∆ ,6MP4.*VEK4W'W "' ( ) = + ⇒ = − = − = ÷ 2 2 2 2 2 2 2 2 2 R 15R OA AH OH AH OA OH 2R 2 4 ⇒ = R 15 AH 2 `a ∆ AKC , ∆ ACH '/ µ A 6 FJFUO'HI%6IR'6 ⇒ ∆ABC CP4F · · ⇒ =ACK ABC c$D@ · = ⇒ o ACO 90 6(#BP4HI%A4@).FJ · · » ⇒ = = 1 ABC AHC sñAC 2 (4HI%O'#BP4HI%A 4@).FJ ] · · =ACK AHC \ ∆ ∆ −AKC ACH(g g)∽ ⇒ = ⇒ = = = = 2 2 AK AC AC 3R 6R 6R 15 AK AC AH AH 15 R 15 15 2 = − = − = R 15 6R 15 R 15 HK AH AK 2 15 10 GHE I#!J)!1!K# 9 !"#%L%% /F$ "44'/ =%?(không kể thời gian giao đề) Z_= #' MN(3,0 điểm) '7484%9#2/ 2 6 9 0x x − + = 7484:%9#2/ 4 3 6 3 4 10 x y y x − = + = 7484%9#2/ 2 6 9 2011x x x − + = − MN%(2,5 điểm) c('MP6M4BGFIJ#4PdBGJIFI 8<41"E,'MD45e$*4I#fN6+MFJ4;= D,,B5<D!41 MN<(2,5 điểm) "#e#B)'443c*g'c*)*gDM1.'4 HI%6IP4c*g,54#B)R'6P4F1"G)Dh,6M ,54)cRFgP4i1"GFDh,6M,54FcR)gP4j1A 4/ 'i)iF "'4@)jFC MN=(2,0 điểm). '"24:6eO'%9#2/ 0 00;<= '4@FJ,6MP4F17&4j4'43@%C4@ #1J4IFJk*jl1"EJ1 WWWWWWWWW.IWWWWWWW OP)*QR)#!S T62UNV2U /F$ W2QN)* 2U MN(3,0 điểm) 4X2Y27!OZ)*1[\)!$ 2 6 9 0x x − + = T Bài giải:"' ' 2 ( 3) 9 0∆ = − − = 0,5 K9#24:/ 6 3 2 x − =− = 0,5 6X2Y2!]7!OZ)*1[\)!$ 4 3 6 (1) 3 4 10 (2) x y y x − = + = T Bài giải:( ,'/<W;0;0< l ⇔ m l ⇔ 0,5 "', /<1;l ⇔ 2 3 1"%4:/ 2 2 3 x y = = 0,5 #X2Y27!OZ)*1[\)!$ 2 6 9 2011x x x − + = − ; T Bài giải:"' ( ) 2 2 6 9 3 3x x x x − + = − = − 0,5 c$D@/ 2 6 9 0 2011 0 2011 3 3x x x x x x − + ≥ ⇒ − ≥ ⇒ ≥ ⇒ − = − \/; 3 2011 3 2011x x ⇔ − = − ⇔ − = 1K9#2,M4: 0,5 MN%(2,5 điểm )/W1#4)0#!^,_N02Q`)*1aEVb)[c2#!^,)*Od#Q`)*1aVb) E!b11S1#Y=*23CJ)!ef)1g##4)0h!2)OP#,i)5j)*T62b1[k)*lNm)*n0)*EQ+2 <h e+ef)1g#Q`)*)OP#5+=h .*23C %To Bài giải: 7&4,O''MD45e$D!4Y< 0,5 \O''MD46M4B0<D!4*D4dBW<D!41 "44''M6M4BGFIJ 30 4x + 4*4dB GJIF 30 4x − 41 0,5 "V4#''%9#2/ 30 30 4 4 4x x + = + − < 0,5 2 ( 4) 30( 4) 30( 4) 4( 4)( 4) 15 16 0 1x x x x x x x ⇔ − + + = + − ⇔ − − = ⇔ =− $ l1g4:W n=eP4 0,5 \,O''MD45e$ lD!41 0,5 MN< (2,5 điểm) [i)VO3)*1[`)pX5S,!42V2U /Tn4'#!'/TTh!0)* 1!q)*!+)*C42rb71N,b)1^2/TeP2VO3)*1[`)pX#s1)!4N1^2ECaht VO3)*eN0)**u#eP2/#s1E1^2CaEhtVO3)*eN0)**u#eP2E/#s11^2 C!K)* 2)!$4XvEC6X4 *2(#E#M) A S O N M I 0,5 4X!K)* 2)!$Ev T \2Fc*Fg@HI%6Ie/ · ¶ MAO SAO = 0,5 \2cF!!i)e/ ¶ ¶ MAO SOA = V# 0,5 "G ,'/ ¶ ¶ SAO SOA = ⇒ ∆ iF)C ⇒ iFi)1%11 6X!K)* 2)!14 *2(#E#M) T \2Fc*Fg@HI%6Ie/ · · MOA NOA = ; 0,5 \2c)!!Fje/ · ¶ MOA OAI = V#< 0,5 "G;,<'/ µ µ IOA IAO = ⇒ ∆ )jFC1%11 MN=(2,0 điểm). 4X\ )*!2] )*N,i)#w47!OZ)*1[\)!$_ % x%, % x%_,x<,&=v T 5 x 6 D B A C I E Bài giải: ⇔ 00 0 0;<= 0,5 ⇔ 0 0W 0<= ⇔ W 0<W 0 \2W 0 ≤ =,54&4*e/W 0< ≤ = ⇔ W< ≤ ≤ 0,5 \26ee ∈ { } 4; 3; 2; 1; 0; 1− − − − "'@4@#6eO','od@$%4:6eQO'K" +/<QW<* QW;*kQW;*WQ=*W Q 1 4X !'14 *2(#EeN0)*1^2EC"25+*24'V2U #(#VO3)*7!M)*2(#1[')*C 2b1Evo# Tvy# CJ)!C 6X #X Bài giải: 7&4]24I6,6MO' #eJj*p4' 43O'FJ,]1 ∆ Jj · DIC 4 e/ · DIC ¶ ¶ $ µ 0 0 1 ( ) 90 : 2 45 2 IBC ICB B C+ = + = = ⇒ DIC∆ ,6MC ⇒ ]l/ 2 c$D@J]%C4@ ,'e'4@Jp 7&4JJp1Y=1q%rsK4W'W,@'4@,6MFJ,Fp '/F J FJ k Wk p F 0Fp Wk0k = / 2 = O,5 !8z$(7()#!{1[\)!6+,#(#!*2Y2Vg2eP2 |26+21'()C(##(#!*2Y2h!(#)bNV8)*eR)#!'V2U 1g2V4C . = 0,5 6X2Y2!]7!OZ)*1[)!$ 4 3 6 (1) 3 4 10 (2) x y y x − = + = T Bài giải:( ,'/<W;0;0<. AB c$D@ ABO ∆ ,6MP4J*VEK4W'W "' ( ) = + ⇒ = − = − = 2 2 2 2 2 2 2 2 2 OA AB OB AB OA OB 2R R 3R (2) "G , ⇒ = 2 AP.AQ 3R ! AHO ∆ ,6MP4.*VEK4W'W "'. = = = 2 2 AK AC AC 3R 6R 6R 15 AK AC AH AH 15 R 15 15 2 = − = − = R 15 6R 15 R 15 HK AH AK 2 15 10 GHE I#!J)!1!K# 9 !"#%L%% /F$ "44'/