!"#$%# &'()*+%$,-+%$. /01 23$1456%738'9 ! " # + = + = $! %&$'()* + #! # = + − − − *! ,-****./0(12+3!&4 # 23+14+:%738'9 %& $&5+6! = 789:+;!<+7=5 /0!> ! ?@6 $&5+6! $! ,-***A*B 8'+6!*C+;!*D8EF8'*(> *! ,G 8F**2'(F*+6!*D&8F$H 5I(8F> 23514$738'9 JK)&&L !"#$%MF8F(2N8A*K- 8&>O*(P$AOK/0D 2Q1-/&7=2N8A>?7R 8F($S/(1(70(*KT2N8A->JUO2N8A8F (*D$ &1(*V*(W$V**(*K/0-$H (X 23,1456.738'9 %&89Y+Z!7F8'[*08AH&+Z!>\]V(V[^W[%7=+Z! +^W%5**V8'!>G_5F8'28F1*(U^%+_O*^7%!>`9 :[_*C+Z!a8')5>Gb5(8'*B _> ! %)8'[W^WZWb*c(F*F89Y>d*8Ae*B 89Y 8D> $! %f · · &'( &)(+ = *! %f[% [_>[7_ +[b <[% !> 2! GgWh5I5i5*V(*B _1*a[^W[%>d*8A7AT*(_/ &*& T*_g>_h8aA5=-> 23.14%6.738'9 %& /02WU #>,AU-*B $'()*6 # j # + − + ; <"!=> 23$1 $:3?3@(ABCDEFGC(H2(I@(ABCDEFGC( ! , *D <$*<1*D "e$" <3 < ?RR"*B 6,5kl<3<m $! # + = = = ⇔ ⇔ + = + = = *! + #! + #! # # # = + − − = + + − − = − 2! , *D < #⇔+!+<!#⇔ # + ! + ! # + ! # # + ! + = = − = = + = = − = = − ⇔ + = − = − − = − = + = − = − − = − = − ?R12*B 5+3! 23+1 ! ?@8nA/0 < < $! do&8F &8'*+6!7+;! J − + ⇔+ << ∆+<! <>+<!j< `'+6!7+;!*DF8'*(∆⇔j<⇔ j ?R7= j +6!7+;!*DF8'*(> *! `'i*+6!&8F$H 5I(8F 51 *D +! ⇔ ⇔ = = ?R8'(F*+6!&8F$H 5I(8F5+3!W+ W ! 2351 G+*V*!/0(2N8A*B 8F+∈pWq! /0( 7R*('5+*V*! k0-1r*V*s&2N8A +-! k0-1r*V*N*V + +-! ,s&8t$ *D < + ⇔+!<+! ⇔ < ⇔ + ! = = − ?R8F(5E*8I(5*V*> 23,1 ! , *Db_b+!⇒Zb⊥_⇒ · j )*+ = _ · j)&+ = +[^5V(V+Z!! k( 8'^Wb(F*89Y 8OT[Z>J [W^WbWZ*c (F*F89YWe*B 89 Y5(8'*B [Z> $! , *D · · &+( &'(= +D*Ke7 D**c*CF*(!> _.O* · · &+( &)(+ = /( · · &'( &)(+ = +8*! *! • do∆[_%7∆[%*D · · · ¼ + ! ')( ,- () (') ./( = = ⇒∆[_% ∽ ∆[%+>! > ) )( )( ) )' )( )' ⇒ = ⇒ = +8*! • , *D[b [Z <Zb +2u`v6< <&7&∆[bZ! [% [Z <Z% +2u`v6< <&7&∆[%Z! k( [b <[% Z% <Zb Z <Zb b ' ' = ÷ _ +[b <[% ! 2! \]_\⊥^%W8&a[Z∩+Z!lwmW[Z∩^%lJm , *D · · 01 (1= +)*_h%\! · · (1 &2= +*c*C**(_%! · · &2 12= +)*_\^g! k( · · 01 12= +! %)N *x*D · · 21 10= +! ,f+!7+!/( ∆_g\ ∽ ∆_\h+>! > 2 1 1 2 '0 1 0 ⇒ = ⇒ = `'_g>_h5=-_\5=->_.O*_(F**(U^%1_\≤wJ⇒7R_\ 5=-O_\wJ>J 3≡ ?RO[W_WZ:_g>_h8aA5=-> 23.1 y2u$8%&/ *D # j + ≥ = +! #≥ # = ⇔ # # # # # # ≤ ⇔ − ≥ − + + +! ,f+!7+!/( 6 # j # + − + ≥ # # − ⇔6 # j # + − + ≥ # ?R_6 # O # + ! # # = = > ⇔ = = ; K=" !L#$%# &'()*+%$,-+%$. /MCE(31NOC ,9 5$EWOzO'9 &8t \&{{ 23$+8'! |EG$'()*[ # − + − + 23++W8'! %& /0< 7 {?@8nA*B **/01*cF.:&a8F {,&a8F &8'*B 8nA/0$HoT 235+8'! {" # # + = − = { }#} #{ } }j 23,+8'! %& }+}!}+5 /0! {%)5(z*D "e$"7=G {,**A*B 8'*D "2R( #{?=A&*B $'()*[ 8aAU->,A8D 23.+#W8'! %&+Z!89OT[^W1 [^5-8'%$1&89Y>,f%O]8&a:%; 7(zD*7=[%7%;[%>0[;*C89Y+Z!a_>\]89:^;*C89 Y+Z!a {%J)[%;5)*FV>d*8A89OT7e*B 89Y&aV )*[%; {%) · · ('4 ()4 = 7~_[^7(z*e #{%)[^>[%[_>[; <<<<<<<<<<<<<<<<<<<<<JV<<<<<<<<<<<<<<<<<<<< =LP< $%-Q/R+%$,-+%$. S3TUCD VW3XC 738' 23$1+8'! [ # − + − + + !>+ ! + ! + !>+ ! − − + − + − + ! + ! − + − + − − + − + W8' W8' 23+1+W8'! {<?@8nA/0< ^A < < < < < < < <?@8nA/0 ^A < ?@8nA8E {6&8F < ó ó+! ó 3 − , 3 7&W *D ?=• ?= − • − ?R&a8F &8'*B 8nA5+3!7+ − 3 − ! W8' W8' W8' W8' 2351+8'! { # # + = − = ó # # # + = − = ó # # j + = = ó # # # + = = ó # # = = ?R"*D"2(-+#3#! {, *D + #! >>+ ! j ∆ = − − − = + = > W8' 6*D "e$" + #! > + #! > − − + = = − − − = = − #{ < }j+! `. + ≥ ! 6+!K <}j+! , *D }$*}+<!+<j! 6+!*D "e$" <+5&a!3 j+R! ?= j j # ⇔ ⇔ = ±= ,R"*B +!*D "5 #3 <# W8' W8' W8' 23,1+8'! }+}!}+p! {, *D~€•<+}!‚ }>+}! }} } }>> +}! •7=G 6+p!5(z*D "e$"7=G {6*D "2-(ó>+}!q ó}q óq óq ?R7=q +p!*D "2-( #{, *D+p!*D "7=G+s& ! 1 + ! + = − = − = − , *D[ } + ! } •[+}! }+}! }} } +! }>>## }# +}#! ƒ ;-(LM O7*„O}#ó # ?R7= # [8aAU-$H W8' W8' W8' W8' W8' W8' W8' 23.1+#W8'! J7@8E {, *D · )(4 j +! · )'4 j +D*FV*C… 89Y! · )(4 · )'4 ;3*c[;2=FD*$Hj ,)*[%;FV89Y89OT W8' [; k(e*B 89Y&aV)*[%;5(8'*B [; {O*($1, *D%;[%7 · )(4 j +! ~[%;7(z*ea% · ()4 , *D · )& j +D*FV*C… 89Y! ~_[^7(z*ea_ O*(+1 , *D,)*[%;FV+*)1! · ('4 · ()4 +%c*C*(%;! , *D · )& j +D*FV*C… 89Y! · &4 j · &4 · &(4 j j ,)*^%;_FV · )& · (4 +*c$c7= · &( !+! , 5a*D[%%;+! ~[%;*ea% · ()4 · (4) · &) · (4 +! ,f+!7+!W/( · )& · &) _ · )& j +%)1! ~_[^7(z*ea_ #{do~[^_7~[;%*D µ [ D**( · )& · )(4 j k( )& )4( ∆ ∆ ∽ )& )4 ) )( ⇒ = > >)& )( ) )4 ⇒ = W8' W8' W8' W8' W8' Q!Q $%Q/R+%$,+%$. /0 '56567 Câu 1: (1,5 điểm) ! }# ! %&" − = − − = >, W$$V"*D" = = Câu 2: (2 điểm) %& }+! #+!>+5 /0! ! ,**A*B 8'+!*D "e$"> ! ,**A*B 8'+!*D "e$" W † 4 > Câu 3: ( 2 điểm) ! |EG$'()* # # # # ) + − = − − + ! ?V89:8‡( 8'[+3!7/&/&7=89: 2> Câu 4 ( 3,5 điểm) %& *8t([^%*D89* &[JW5-8'_cˆ(F*8&aJ%+_Ozc7=JW%!> J*V(7(zD**B _51***a[^W[%5I5i567‰> ! %)H[6_‰5)*FV7*8AeZ*B 89Y&aV)* [6_‰> ! %)H^6>^[^J>^_ #! %)HZJ ⊥ 6‰> ! %)HO_ 8S1J%_6_‰Oz8S> Câu 5 (1 điểm) ,AU-*B $'()* # ) + = + − + + 7=•> !Y<"!=> Câu 1: (1,5 điểm) ! }# $*+<#! ⇒ 3 , > ! J" − = − − = *D" = = ⇔ # # # # # − = − − = − − = = − ⇔ ⇔ ⇔ − = = = = > Câu 2: (2 điểm) %& }+! #+!>+5 /0! ! [ ] Š + !∆ = − + <+ #!<} 6+!*D"e$" ⇔ Š ∆ • ⇔ <}• ⇔ q< ?R7=q<+!*D"e$"> #! ?=q<>,s&")*?<s *D +!3 #> ⇔ + ! < ⇔ +! }+ #! ⇔ } 6, *D +Oz,_`\!3 <#+,_`\!> ?R7=<#+!*D"e$"† 4 > Câu 3: ( 2 điểm) 1) |EG$'()* # # # # # # + # ! + # ! ) + − + − = − = − − + − + # # # # + # ! + #! # # + # ! + # ! + − + − − = − = + − − − + − + + # ! + #! + # #!+ # #! #+ − − = + + − + − + = > ! 689:*I7V*D2a2€ $> 2Š8‡( 8'[+3! ⇔ >$ ⇔ $> 2Š /&/&7=89:2 < ⇔ <> ?R*I7V52€<> Câu 4 ( 3,5 điểm) 1) do)*[6_‰*D · · j8) 9)= = +,s&,! ⇒ · · 8) 9)+ = ⇒ )*[6_‰FV> ,eZ*B 89Y&aV)*[6_‰5(8'*B [_ 2) do ∆ ^6_7 ∆ ^J[*D · · j&8 &:)= = +!3 · · 8& :&)= +*(D*^! ⇒ ∆ ^6_ : ∆ ^J[+>! ⇒ &8 & &: &) = ⇒ ^6>^[^J>^_ 3) · j): = +! ⇒ J(F*89Y89OT[_ ⇒ [W6WJW_W‰*c(F*89YZ> · · 8): 9):= +7 *[^%8t(W[J589* &1*x589e*! ⇒ ¼ ¼ 8: 9:= ⇒ 6J‰J ⇒ J(F*89(N**B 6‰+! Z6ZJ+*c$OT! ⇒ Z(F*89(N**B 6‰+! ,f+!7+! ⇒ ZJ589(N**B 6‰ ⇒ ZJ ⊥ 6‰> !k [^_ k %[_ k [^% ⇔ [^>_6 [%>_‰ ^%>[J ⇔ ^%>_6 ^%>_‰ ^%>[J+7[^[%^%! ⇔ ^%+_6_‰! ^%>[J ⇒ _6_‰[J>?[JOz8S 1_6_‰Oz8S> Câu 5 (1 điểm). ?=•W *D # # + ! + ! + ! > + ! + ! + ! ) ) + + = + − + = − + + − + + + − + = − + + + + − = − + + ≥ + − = ⇒ = ⇔ ⇔ = − = [...]... 2)2014 + 2015 = (1)2014 + 2015 = 1 + 2015 = 2016 S GIO DC V O TO TY NINH Kè THI TUYN SINH VO LP 10 NM HC 2014 2015 Ngy thi : 21 thỏng 6 nm 2014 Mụn thi : TON (Khụng chuyờn) Thi gian : 120 phỳt (Khụng k thi gian giao ) CHNH THC ( thi cú 01 trang, thớ sinh khụng phai chộp vo giy thi) Cõu 1 : (1im) Thc hin cỏc phep tinh ( a) A = 2 5 ) ( 2 + 5) b) B = 2 (... HN, nờn H l trung im cua EN Suy ra HK l ng trung bỡnh cua tam giỏc EAF Võy HK // AF Võy ED // HK // AF S GIO DC V O TO KHNH HO K THI TUYN SINH VO LP 10 THPT CHUYấN NM HC 2014 2015 MễN THI: TON (KHễNG CHUYấN) Ngy thi: 20/6/2014 (Thi gian : 120 phỳt khụng k thi gian giao ) THI CHNH THC Bi 1: (2,00 im) 1) Khụng dung mỏy tinh cõm tay, tinh giỏ tri biu thc: A = 1 8 10 2 +1 2 5 a a a +1 + 2) Rut gon...S GIO DC V O TO BèNH NH D B K THI TUYN VO LP 10 TRUNG HC PH THễNG NM HC 2014 2015 Mụn thi: TON Ngy thi: 28/6/2014 Thi gian lm bi: 120 phỳt (khụng k thi gian phỏt ) Bi 1: (2,5 im) a) Gii phng trỡnh: 3x 5 = x + 1 b) Gii phng trỡnh: x 2 + x 6 = 0 x 2 y = 8 c) Gii h phng trỡnh: x + y = 1... T (1) v (2) suy ra: AM = R 2 AOM vuụng tai O M l im chinh gia cung AB S GIO DC V O TO QUNG NGI K THI TUYN SINH VO LP 10 THPT NM HC 2014-2015 MễN : TON (khụng chuyờn) Ngy thi: 19/6/2014 Thi gian lm bi: 120 phỳt (khụng k thi gian giao ) CHNH THC Bi 1: (1,5 im) a/ Tinh: 2 25 + 3 4 b/ Xỏc inh a v b ụ thi hm sụ y = ax + b i qua im A(1; 2) v im B(3; 4) c/ Rut gon biu thc A = x x +2 + x+4 : vi x... GIO DC V O TO TP. NNG CHNH THC K THI TUYN SINH LP 10 THPT NM HC 2014-2015 MễN: TON Thi gian lm bi: 120 phỳt Bi 1: (1,5 im) 1) Tinh giỏ tri cua biu thc A = 9 4 Rut gon biu thc P = x 2 2x 2 + , vi x > 0, x 2 x2 2 x+x 2 Bi 2: (1,0 im) 3 x + 4 y = 5 Gii h phng trỡnh 6 x + 7 y = 8 Bi 3: (2,0 im) Cho hm sụ y = x2 co ụ thi (P) v hm sụ y = 4x + m co ụ thi (dm) 1)V ụ thi (P) 2)Tỡm tõt c cỏc giỏ tri cua... thc A = (4x5 + 4x4 5x3 + 5x 2)2014 + 2015 Tinh giỏ tri cua biu thc A khi x = 1 2 2 1 2 +1 - HấT Giỏm thi coi thi khụng gii thich gỡ thờm GI í BI GII TON VO 10 KHễNG CHUYấN Lấ KHIấT QUNG NGI Bi 1: a/ Tinh: 2 25 + 3 4 = 10 + 6 = 16 b/ ụ thi hm sụ y = ax + b i qua A(1; 2) nờn a + b = 2, v B(3; 4) nờn 3a b = 4 Suy ra a = 3, b = 5 Võy (d): y = 3x + 5 x 2 x+4 1 x +2 :... - HấT S GIAO DC O TO NINH THUN K THI TUYN SINH VO LP 10 THPT NM HC 2014 2015 Khoa ngy: 23 6 2014 Mụn thi: TON Thi gian lm bi: 120 phut -Bi 1: (2,0 im) a) Gii phng trỡnh bõc hai: x2 2x 2 = 0 3 x + y = 2 b) Gii h phng trỡnh bõc nhõt hai õn: 2( x y ) 5 x = 2 Bi 2: (2,0 im) Cho hm sụ: y = 2x 5 co ụ thi l ng thng (d) a) Goi A, B lõn lt... + ab 2 Cụng (1) (2) (3) vờ theo vờ Q 2(a + b + c) = 4 2 Khi a = b = c = thỡ Q = 4 võy giỏ tri ln nhõt cua Q l 4 3 0,25 0,25 S GIO DC V O TO K THI TUYN SINH VO LP 10 THPT NM HC 20142015 PHU TH Mụn : TON Thi gian lm bi: 120 phut, khụng k thi gian giao ờ thi cú 01 trang CHNH THC Cõu 1 (1,5im) a) Trong cỏc phng trỡnh di õy, nhng phng trỡnh no l phng trỡnh bõc 2: x 2 + 3x + 2 = 0; 3x 2 + 4 = 0 2x... chuyn xong mụt na sụ lng thc, thc phõm; sau o ngi th hai chuyn hờt sụ con lai lờn tu thỡ thi gian ngi th hai hon thnh lõu hn ngi th nhõt l 3 gi Nờu c hai cung lm chung thỡ thi gian chuyn hờt sụ lng thc, thc phõm lờn tu l 20 7 gi Hoi nờu lm riờng mụt mỡnh thỡ mụi ngi chuyn hờt sụ lng thc, thc phõm o lờn tu trong thi gian bao lõu? Bi 4: (3,5 im) Cho na ng tron tõm O, ng kinh AB = 2R Goi M l im chinh gia... 10 : (1 im) Cho t giỏc ABCD nụi tờp ng tron tõm O, bỏn kinh bng a Biờt AC vuụng goc vi BD Tinh AB2 + CD 2 theo a - HT Giỏm th khụng gii thớch gỡ thờm Ho v tờn thi sinh : Sụ bỏo danh : Ch ký cua giỏm thi 1: Ch ký cua giỏm thi 2 : BI GII Cõu 1 : (1im) Thc hin cỏc phep tinh ( ) ( 2 + 5 ) = 2 ( 5 ) = 4 5 = 1 2 ( 50 3 2 ) = 100 3.2 = 10 6 = 4 a) A = 2 5 b) B = 2 2 Cõu 2 : . &2= +*c*C**(_%! · · &2 12= +)*_^g! k( · · 01 12= +! %)N *x*D · · 21 10= +! ,f+!7+!/( ∆_g ∽ ∆_h+>! > 2 1 1 2 '0 1 0 ⇒ = ⇒ = `'_g>_h5=-_5=->_.O*_(F**(U^%1_≤wJ⇒7R_ 5=-O_wJ>J