CNG ễN TP HK MễN TON LP 10 NM HC 2013-2014 TRNG THPT THI PHIấN A/ Chng trỡnh c bn: I/ i s: - Mnh - Tp hp, cỏc phộp toỏn trờn hp, cỏc hp s - Hm s - s bin thiờn, tớnh chn l ca hm s; hm s bc nht, hm s bc hai - Phng trỡnh; gii v bin lun phng trỡnh bc nht; phng trỡnh bc hai; ng dng ca nh lý Viet; phng trỡnh quy v phng trỡnh bc nhõt, bc hai - H phng trỡnh bc nht nhiu n II/ Hỡnh hc: - Vect; tng v hiu ca vecto; phộp nhõn mt vecto vi mt s - H trc ta - Giỏ tr lng giỏc ca mt gúc 00 1800 - Tớch vụ hng ca hai vecto B/ Chng trỡnh nõng cao: I/ i s: - Mnh ; mnh cha bin - p dng mnh vo suy lun toỏn hc - Tp hp, cỏc phộp toỏn trờn hp, cỏc hp s - Hm s - s bin thiờn, tớnh chn l ca hm s; hm s bc nht, hm s bc hai - Phng trỡnh; gii v bin lun phng trỡnh bc nht; gii v bin lun phng trỡnh bc hai; ng dng ca nh lý Viet; phng trỡnh quy v phng trỡnh bc nhõt, bc hai - H phng trỡnh bc nht nhiu n - H phng trỡnh bc hai hai n - Bt ng thc II/ Hỡnh hc: - Vecto; tng v hiu ca vecto; phộp nhõn mt vecto vi mt s - H trc ta - Giỏ tr lng giỏc ca mt gúc 00 1800 - Tớch vụ hng ca hai vecto - H thc lng tam giỏc ( gii tam giỏc ) 1: I PHN CHUNG CHO TT C TH SINH ( im ) Cõu I ( 2,0 ) : Cho cỏc hp: A = { x Ă | x 1} , B = { x Ă | < x < 2} Dựng kớ hiu on, khong, na khong vit li cỏc hp trờn v xỏc nh A B; A B; A \ B; CĂ B Cõu II ( 2,0 ): 1) Lp bng bin thiờn v v th ca hm s: y = x + x + 2) Gii v bin lun phng trỡnh: (m x 1)m = x Cõu III ( 3,0 ): 1) Cho t giỏc ABCD Gi M, N ln lt l trung im AD, BC Chng minh: uuur AB + uuur DC uuuu r = MN v uuur AC + uuur DB uuuu r = MN 2) Trong mt phng ta Oxy cho: A(2;5), B(1;3) v C(5;5) a) Chng minh A, B, C l ba nh ca mt tam giỏc b) Tỡm to im M cho: uuuu r MA uuur MB = uuur BC II PHN RIấNG ( im ) Thớ sinh ch c lm mt hai phn ( phn A hoc phn B ) A.Theo chng trỡnh chun : Cõu IV.a ( 1,0 im ) : Tỡm m phng trỡnh: nghim x1 , x2 tha: x1 + x1 x2 + x2 = x 2(m + 1) x + m 5m = cú Cõu V.a ( 2,0 im ) : Gii phng trỡnh: a) x + 3x = x + ; b) ( x 4)( x + 6) + x + x + + = B.Theo chng trỡnh nõng cao : Cõu IV.b ( 1,0 im ) : Cho phng trỡnh: mx + 2(m 2) x + m = (1) nh m phng trỡnh (1) cú hai nghim Cõu V.b ( 2,0 im ) : x1 , x2 tha: x1 x2 + = x2 x1 hai a Gii h phng trỡnh: x + y + xy = 2 x + y + xy = b tỡm m h phng trỡnh sau cú nghim mx + (2m + 1) y = 3m (2m + 1) x + my = 3m + 2: I PHN CHUNG CHO TT C TH SINH ( im ) Cõu I: (2) Cho hm s y = x x + a/ V th (P) ca hm s b/ Tỡm to giao im ca (P) vi ng thng y=3x-3 Cõu II: (2) 1/ Cho cỏc hp: C = { x R | x 4} A = { x R | x < 3} B = { x R |1 < x 5} A B, B C , A \ C Tỡm 2/ Chng minh rng vi ba s a, b, c dng ta cú: + a ữ + b ữ + c ữ abc b c a a b c Cõu III: (3) Trong mt phng to Oxy cho A( ;1 ) B( 1; 4) C(2 ; -1) a/ Chng minh rng tam giỏc ABC l tam giỏc vuụng b/ Tỡm to tõm I v bỏn kớnh R ca ng trũn ngoi tip tam giỏc ABC c/ Tỡm to im H l hỡnh chiu vuụng gúc ca A trờn BC II PHN RIấNG ( im ) Thớ sinh ch c lm mt hai phn ( phn A hoc phn B ) A.Theo chng trỡnh chun : Cõu IV.a (1) Cho phng trỡnh ( m + 1) x + 2mx + m = 2 Tỡm giỏ tr ca m phng trỡnh cú hai nghim phõn bit x1 , x cho x + x = Cõu V.a (2) : Gii cỏc phng trỡnh sau: a/ x 20 = x 20 b/ (x2 + 3x +2)(x2+7x +12)=24 B.Theo chng trỡnh nõng cao : Cõu IV.b ( 1,0 im ) : Gii v bin lun phng trỡnh sau (vi m l tham s) ( m + 3) x + m mx = x m x x+3 Cõu V.b ( 2,0 im ) : 1/ Gii phng trỡnh: x + x + - = 3x 2/ Cho phng trỡnh: 4x2-5x+1=0 cú hai nghim l x , x Khụng gii phng trỡnh tớnh giỏ tr ca biu thc sau: A = x1 + x2 3: I PHN CHUNG CHO TT C TH SINH ( im ) Cõu I ( 2,0 ) Cho hm s : y = ax + x a0 a Xỏc nh hm s bit th hm s i qua A(1;2) b Lp bng bin thiờn v v th hm s va tỡm c Cõu II ( 2,0 ) Gii cỏc phng trỡnh sau: a/ b/ 3x x = x + x +3= x2 x2 Cõu III ( 2,0 ) 1/ Trong mt phng ta Oxy cho uuu r r r uuur r r OA = i + j ; B(-4;-5) ; OC = 4i j a/ Chng minh ba im A, B, C l nh ca tam giỏc b/ Tỡm D cho t giỏc ADCB l hỡnh bỡnh hnh 2/ Cho tam giỏc ABC cú trng tõm G Gi D v E l cỏc im xỏc nh bi uuu r uuur AE = AC Biu din uuur uuur uuu r uuur AG, DG theo AB va AC uuur uuu r AD = AB , Cõu IVuuu:r u(1,0) Cho tam giỏc ABC vuụng ti A cú AC = 10 v AB = 22 Tớnh tớch vụ uu r hng CA.CB II PHN RIấNG ( im ) Thớ sinh ch c lm mt hai phn ( phn A hoc phn B ) A.Theo chng trỡnh chun : Cõu V.a ( 1,0 ) : Gii phng trỡnh | x | = 3x2 x Cõu VI.a ( 2,0 ) : Cho phng trỡnh mx + 2(m 4) x + m + = a/ Tỡm m phng trỡnh cú hai nghim phõn bit b/ Tỡm m phng trỡnh cú hai nghim x1 , x2 tho x1 , x2 x1 x2 = B.Theo chng trỡnh nõng cao : Cõu V.b ( 1,0 ) : Tỡm giỏ tr ca tham s m h phng trỡnh sau cú nghim (m 2) x + 5y = m (m + 3) x + (m + 3) y = 2m Cõu VI.b ( 2,0 ) : Cho tam giỏc ABC cú a = 5, b = 6, c = Tớnh: a) Din tớch S ca tam giỏc b) Tớnh cỏc bỏn kớnh R, r c) Tớnh cỏc ng cao ha, hb, hc 4: I PHN CHUNG CHO TT C TH SINH ( im ) Cõu I ( 2,0 ) Cho cỏc hp: { } A = x R |( x2 + x + 6) ( x2 4) = B = { x N |2 x 8} ; C = { x + 1| x Z , x 4} a/ Hóy vit li cỏc hp A, B, C di dng lit kờ cỏc phn t b/ Tỡm A B, A B, B \ C , ( A C ) \ B Cõu II ( 2,0 ) Cho hm s : y = mx -2(m-1)x+3 (m 0) a Xỏc nh hm s bit th ca nú cú trc i xng x = b Kho sỏt v v th hm s va tỡm c c Tỡm ta giao im ca parabol trờn v ng thng y = x + Cõu III ( 3,0 ) 1/Cho tam giỏc ABC trng tõm G Gi D v E l cỏc im xỏc nh bi uuur uuu r uuur uuur AD = AB, AE = AC Biu din uuur DE v uuur DG theo uuur AB v uuur AC 2/ Trong mt phng ta cho A(-1; 4) ; B(1; 1) v C( -4; -2) a Chng minh ba im A, B, C to thnh mt tam giỏc b Tỡm ta im D cho ABCD l hỡnh bỡnh hnh II PHN RIấNG ( im ) Thớ sinh ch c lm mt hai phn ( phn A hoc phn B ) A.Theo chng trỡnh chun : Cõu IV.a ( 1,0 ) : Gii phng trỡnh: x + = x +3 x x Cõu V.a ( 2,0 ) : cho hm s f ( x ) = mx -2(m+1)x+m-5 a Tỡm m phng trỡnh f ( x) = cú nghim b Vi iu kin cú nghim nh trờn, tỡm giỏ tr m hai nghim ca 2 phng trỡnh tha x + x1 = B.Theo chng trỡnh nõng cao : Cõu IV.b ( 1,0 ) : Gii pt: x + = 2x + Cõu V.b ( 2,0 ) : Cho phng trỡnh x2 ( m 2) x + m2 = a Tỡm m ptrỡnh cú nghim x = Tớnh nghim cũn li b Tỡm m ptrỡnh cú hai nghim x1, x2 tho : x12+x22 = 26 Cõu V.b ( 2,0 ) : 1/ Gii h phng trỡnh x 2x = y y 2y = x 2/ Cho phng trỡnh ( n x) : x4 - 2mx2 + m2 = Tỡm m phng trỡnh cú ỳng nghim phõn bit 6: I PHN CHUNG CHO TT C TH SINH ( im ) Cõu I ( 2,0 ) :Cho cỏc hp sau : A = { x Ơ * / x 4} { x  / -2 x < 4} B = { x Ă / 2x( 3x2 2x 1) = 0} C= a) Hóy vit li cỏc hp di dng lit kờ cỏc phn t b) Hóy xỏc nh cỏc hp sau : A C, A B, C\B, (C\A) B Cõu II ( 2,0 ) f ( x ) = ax + bx + c a Xỏc nh hm s bit th hm s cú nh S(2; -1) v i qua im M(1; 0) b Lp bng bin thiờn v v th hm s va tỡm c Cõu III ( 3,0 ) 1/ Cho tam giỏc ABC Gi M l mt im trờn on BC, cho MB= 2MC Chng uuuu r uuu r uuur 3 minh rng : AM = AB + AC 2/ Cho im A(1,2), B(2, 6), C(4, 4) a/ Chng minh A, B,C khụng thng hng b/ Tỡm to im N cho B l trung im ca on AN c/ Tỡm to cỏc iờm H, Q, K cho C l trng tõm ca tam giỏc ABH, B l trng tõm ca tam giỏc ACQ, A l trng tõm ca tam giỏc BCK II PHN RIấNG ( im ) Thớ sinh ch c lm mt hai phn ( phn A hoc phn B ) A.Theo chng trỡnh chun : Cõu IV.a ( 1,0 ) : Tỡm m phng trỡnh cú nghim vi mi x R : m x + = x + 3m Cõu V.a ( 2,0 ): Gii cỏc phng trỡnh sau: 1/ 2x +1 = 2x 2/ x x 2x = x x + ( x + 1)( x 3) B.Theo chng trỡnh nõng cao : Cõu IV.b ( 1,0 ) : Cho phng trỡnh (m -1)x2 - 2mx + m + = Tỡm m phng trỡnh cú hai nghim x1, x2 tha h thc 5(x1 + x2) 4x1 x2 - = x + + y = 11 Cõu V.b ( 1,0 ) : Gii h phng trỡnh: = x + y Cõu VI.(1,0) Cho gúc x vi cosx = P = 2sin2x + 3cos2x Tớnh giỏ tr ca biu thc: 7: I PHN CHUNG CHO TT C TH SINH ( im ) Cõu I ( 2,0 ) 1/ Cho A = (0;2] v B = [1;4) Tỡm CR(A B) v CR(A B) 2/ Xỏc nh cỏc A v B bit rng : A B = {3,6,9} ; A\B = {1,5,7,8} ; B\A = {2,10} Cõu II ( 2,0 ) Cho hm s y = x2 + mx -3 (1) a/ Tỡm m th hm s (1) ct trc Ox ti im cú honh bng b/ Lp bng bin thiờn v v th hm s (P) ca hm s (1) m = -3 c/ Tỡm to giao im ca th (P) vi ng thng (d) : y = 2x + Cõu III ( 3,0 ) 1/ Cho tam giỏc ABC , gi M l trung im ca AB, N l mt im trờn AC cho NC=2NA, gi K l trung im ca MN uuur uuur uuur CMR: AK= AB + AC 2/ Cho im A(1,2), B(2, 6), C(4, 4) Tỡm to im T cho im A v T i xng qua B, qua C 3/ Chng minh rng: ( tan + cot ) ( tan cot ) = vi bt kỡ II PHN RIấNG ( im ) Thớ sinh ch c lm mt hai phn ( phn A hoc phn B ) A.Theo chng trỡnh chun : Cõu IV.a ( 1,0 ) : Gii phng trỡnh: 5x + = x Cõu V.a ( 2,0 ) : Cho phng trỡnh : x 2(5 + m) x + 5m = a Tỡm m phng trỡnh cú nghim b Tỡm m phng trỡnh cú mt nghim gp ln nghim B.Theo chng trỡnh nõng cao : x2 x x = Cõu IV.b ( 1,0 ) : Gii phng trỡnh: x x Cõu V.b ( 2,0 ) : Cho h phng trỡnh : mx y = x + my = a/ Tỡm m h cú nghim nhõt Gi nghim ca h phng trỡnh l (x, y) Tỡm cỏc giỏ tr ca m x + y = -1 b/ Tỡm ng thc liờn h gia x v y khụng ph thuc vo m 8: I PHN CHUNG CHO TT C TH SINH ( im ) Cõu I ( 1,0 ) : Cho cỏc hp: A = { x R | x < 3} B = { x R |1 < x 5} C = { x R | x 4} A B, B C , A \ C Tỡm Cõu II ( 2,0 ) Cho hm s y = x x + a/ V th (P) ca hm s b/ Tỡm to giao im ca (P) vi ng thng y=3x-3 Cõu III ( 3,0 ) 1/ Gi E, F ln lt l trung im cỏc cnh AD v BC ca t giỏc ABCD Chng minh rng: AB+ DC = EF 2/ Cho ba im A(1; 5), B(3; 1), C(1; 0) a/ Chng minh ba im A, B, C l ba nh ca mt tam giỏc b/ Tỡm ta im M cho MA MB = II PHN RIấNG ( im ) Thớ sinh ch c lm mt hai phn ( phn A hoc phn B ) A.Theo chng trỡnh chun : Cõu IV.a ( 1,0 ) : Gii phng trỡnh: C x +8 = +1 x2 x+2 x Cõu V.a ( 2,0 ) : Cho phng trỡnh x 2(m + 1) x + 2m + 10 = (1) 1/ Gii phng trỡnh vi m = -6 2/ Tỡm m phng trỡnh (1) cú hai nghim A = x + x + 10 x1 x2 2 B.Theo chng trỡnh nõng cao : Cõu IV.b ( 1,0 ) : x1 , x2 Tỡm GTNN ca biu thc Gii h phng trỡnh: x + y + xy = 11 2 x y + y x = 30 Cõu V.b ( 2,0 ): 1/ Gii phng trỡnh x + 14 x + = x + 2/ Tỡm m ng thng y= -3x-2 ct parabol y = x x + m ti im phõn bit 9: I PHN CHUNG CHO TT C TH SINH ( im ) Cõu I (1,0 ) : 1/ Xột tớnh ỳng, sai ca mnh sau : ca nú P = " x R : x 3x 0" Lp mnh ph nh Cõu II ( 2,0 ) 1/ Lp bng bin thiờn v v th hm s 2/ Da vo th hm s |x x |= m y=|x x | y = x2 x bin lun theo m s nghim ca phng trỡnh Cõu III ( 3,0 ) 1/ Cho tam giỏc u ABC cnh 2a, gi M l im nm trờn cnh BC cho MB=2MC, N l trung im ca AC a b uuuu r uuur uuur CMR : MN = AB AC uuur uuur uuuu r uuur uuuu r Phõn tớch AM theo vộct AB,AC Tinh AM.MC 2/ Trong mp oxy cho im A(1;1), B(-2;3) Xỏc nh ta im D cho O l trng tõm tam giỏc ABD Cõu IV : (1,0) Cho a>0; b>0 Chng minh rng a b + a + b ng thc xy b a no? II PHN RIấNG ( im ) Thớ sinh ch c lm mt hai phn ( phn A hoc phn B ) A.Theo chng trỡnh chun : Cõu V.a ( 1,0 ) : Tỡm xỏc nh ca hm s: Cõu VI.a ( 2,0 ) : Cho phng trỡnh x +1 x 4x + 4x (m + 2)x 2(m 1)x + m = a/ Gii v bin lun phng trỡnh b/ Xỏc nh m pt (1) cú ỳng nghim dng B.Theo chng trỡnh nõng cao : y= Cõu V.b ( 1,0 ) : Gii h phng trỡnh: Cõu VI.b ( 2,0 ): Cho phng trỡnh: x + y = 130 xy x y = 47 x + 2mx + m = a/ Xỏc nh m phng trỡnh cú hai nghim b/ Gi A= l nghim ca phng trỡnh, tỡm giỏ tr ln nht ca biu thc : x1 x2 + x1 + x2 x1 , x2 10: I PHN CHUNG CHO TT C TH SINH ( im ) Cõu I ( 1,0 ) : Tỡm xỏc nh ca hm s: a) y= x2 4x b) y= x +1 x 2x + Cõu II ( 3,0 ) 1/ Cho parabol (P): y = ax2 + bx + c ( a ) a/Tỡm a, b, c bit rng (P) i qua im A(0;3) v cú nh I(2; -1) b/ Lp bng bin thiờn v v th hm s tỡm c cõu a 2/ Tỡm m phng trỡnh x12 + x 22 = x + 5x + 3m = cú hai nghim phõn bit x1 , x tha Cõu III ( 3,0 ) 1/Trong mt phng ta Oxy cho im A ( 3; 1) , B ( 2; ) , C ( 5;3) a Tỡm M cho C l trng tõm tam giỏc ABM b Tỡm N cho tam giỏc ABN vuụng cõn ti N 2/Cho tam giỏc ABC, gi I l trung im BC Kộo di CA mt on AN = AC, kộo di BA mt on uuur r uuur uuu AK = AB + AC ( AM = ) AB Gi K l im tha món: uuur uuur r KM + KN = Chng minh: II PHN RIấNG ( im ) Thớ sinh ch c lm mt hai phn ( phn A hoc phn B ) A.Theo chng trỡnh chun : Cõu IV.a ( 1,0 ) : Gii phng trỡnh: x + = x2 + 5x + = 600 , AC=1, AB=3 Trờn cnh AB ly im Cõu V.a ( 1,0 ) : Cho tam giỏc ABC cú A uuur uuur D cho BD=1 Gi E l trung im CD Tớnh AE.BC Cõu VI.a ( 1,0 ): Tỡm GTNN ca hm s B.Theo chng trỡnh nõng cao : f (x) = x + 2x vi mi x> Cõu IV.b ( 1,0 ) : Gii phng trỡnh : x + 15 x + x + x + = Cõu V.b ( 1,0 ): Gii h phng trỡnh : x -2y = 2x + y 2 y -2x =2y + x Cõu VI.b (1,0): Cho tam giỏc cõn ABC cú AB = AC = a v uuur uuu r uuu r uuu r uuur uuu r biu thc: T = AB.CB + CB.CA + AC.BA theo a ã BAC = 1200 Tớnh giỏ tr ca TRNG THPT THI PHIấN KIM TRA HC K I NM HC 2011 2012 MễN THI: TON - KHI LP 10 Thi gian: 90 phỳt (khụng k thi gian giao ) CHNH THC I/ PHN BT BUC (7,0 im) Cõu I (1,0 im): Cho cỏc hp: A = ( 5; ) ; A B ; A\ B ; B \ A Cõu II (1,0 im): Gii phng trỡnh sau: Cõu III (2,0 im): Cho hm s: B = [ 2;9 ) Xỏc nh cỏc hp: A B ; 3x + = x + y = f ( x) = x x cú th l parabol (P) 1/ Lp bng bin thiờn v v parabol (P) 2/ Tỡm ta giao im ca (P) v ng thng (d): y = x +1 3/ Tỡm giỏ tr ln nht, giỏ tr nh nht ca hm s trờn [ 0;3] Cõu IV (3,0 im): 1/ Trong h trc Oxy, cho A(1;2); B(4;1); C(-2;-3) a/ CM: A, B, C l nh ca mt tam giỏc Tỡm ta trng tõm G ca b/ Tỡm im M thuc trc Ox cho uuur uuur MA + MB nh nht uuuu r 2/ Cho tam giỏc ABC cú G l trng tõm, ly im M tha AM uuur uuur AN = AC CMR: im M, G, N thng hng uuu r = AB , II/ PHN T CHN (3,0 im) (Hc sinh ch c chn mt hai phn: phn A hoc phn B) A/ Theo chng trỡnh chun: Cõu Va (2,0 im): 1/ Gii phng trỡnh sau: 2/ Cho phng trỡnh: 5x = x x ( m 1) x + m 3m = a/ nh m phng trỡnh trờn cú nghim ABC (m l tham s) v im N tha b/ Gi x1 ; x2 l hai nghim ca phng trỡnh trờn nh m ( x1 + x2 ) + x1 x2 + = Cõu VIa (1,0 im): Cho phng trỡnh: m ( x ) m nh m phng trỡnh trờn cú nghim x ( 1; + ) = x 3, ú m l tham s Xỏc B/ Theo chng trỡnh nõng cao: Cõu Vb (2,0 im): 1/ Gii h phng trỡnh sau: 2/ Cho phng trỡnh: x + y = 10 2 x + y = 58 mx ( m + 3) x + m = (m l tham s) a/ nh m phng trỡnh cú nghim phõn bit x1 ; x2 b/ nh m phng trỡnh cú nghim phõn bit x1 ; x2 tha: x12 + x22 = 10 Cõu VIb (1,0 im): Cho hai ng thng d1 : ( m n ) x + y = ; d : ( m n ) x + my = n (vi m, n l tham s v m2 + n ) Tỡm iu kin ca m v n ng thng trờn ct ti mt im thuc trc honh 2 TRNG THPT THI PHIấN KIM TRA HC K I NM HC 2012 2013 MễN THI: TON - KHI LP 10 Thi gian: 90 phỳt (khụng k thi gian giao ) CHNH THC I/ PHN BT BUC (7,0 im) Cõu I (1,0 im): Tỡm hp A v B bit A B = ( -;5 , A \ B = ( ; , B \ A = 3;5 Cõu II (2,0 im): Cho Parabol ( Pm) : y = x 2mx ( m tham s ) 1/ Tỡm m ( Pm) nhn ng thng x = lm trc i xng Lp bng bin thiờn v v ( Pm) vi m va tỡm c 2/ Tỡm m ng thng d : y = - x - ct ( Pm) ti hai im phõn bit A, B cho OA vuụng gúc vi OB Cõu III (1,0 im): Gii phng trỡnh sau: x x x + 3x + 10 + = x x + x2 Cõu IV (3,0 im): 1/ Trong h trc ta Oxy, cho A(1;-1); B(5;-3); C(2;0) a/ Chng minh A, B, C l nh ca mt tam giỏc Tam giỏc ABC l tam giỏc gỡ? b/ CH l ng cao h t C ca tam giỏc ABC, tỡm ta im H 3sin cos 2/ Cho tan = Tớnh giỏ tr biu thc: M = sin + cos II/ PHN T CHN (3,0 im) (Hc sinh ch c chn mt hai phn: phn A hoc phn B) A/ Chng trỡnh chun: Cõu V.a (1,0 im): Gii phng trỡnh: 3x + + = x Cõu VI.a (1,0 im): Cho phng trỡnh: x + ( m 1) x + m = (m l tham s) Gi x1; x2 l hai nghim ca phng trỡnh trờn Tỡm m x12 + x22 t giỏ tr nh nht Cõu VII.a (1,0 im): Tỡm xỏc nh ca cỏc hm s sau : a/ y = x 14 3x x 2x b/ y = x + + x B/ Chng trỡnh nõng cao: x + y = Cõu V.b (1,0 im): Cho h phng trỡnh sau: x + y = 10m + Vi giỏ tr no ca m thỡ tớch hai nghim x.y t giỏ tr ln nht Cõu VI.b (1,0 im): Cho phng trỡnh ( m + 1) x ( m 1) x + m = Tỡm m phng trỡnh cú nghim phõn bit x1; x2 tha: x1 = 3x2 Cõu VII.b (1,0 im): Gii phng trỡnh : x x + x = Ht [...]... phn B ) A.Theo chng trỡnh chun : Cõu IV.a ( 1,0 ) : Gii phng trỡnh: C x +8 1 4 = 2 +1 x2 x+2 x 4 Cõu V.a ( 2,0 ) : Cho phng trỡnh x 2 2(m + 1) x + 2m + 10 = 0 (1) 1/ Gii phng trỡnh vi m = -6 2/ Tỡm m phng trỡnh (1) cú hai nghim A = x + x + 10 x1 x2 2 1 2 2 B.Theo chng trỡnh nõng cao : Cõu IV.b ( 1,0 ) : x1 , x2 Tỡm GTNN ca biu thc Gii h phng trỡnh: x + y + xy = 11 2 2 x y + y x = 30 Cõu V.b... l tham s Xỏc B/ Theo chng trỡnh nõng cao: Cõu Vb (2,0 im): 1/ Gii h phng trỡnh sau: 2/ Cho phng trỡnh: x + y = 10 2 2 x + y = 58 mx 2 2 ( m + 3) x + m 2 = 0 (m l tham s) a/ nh m phng trỡnh cú 2 nghim phõn bit x1 ; x2 b/ nh m phng trỡnh cú 2 nghim phõn bit x1 ; x2 tha: x12 + x22 = 10 Cõu VIb (1,0 im): Cho hai ng thng d1 : ( m n ) x + y = 1 ; d 2 : ( m n ) x + my = n (vi m, n l tham s v m2... phng trỡnh: x 2 + y 2 = 130 xy x y = 47 2 x 2 + 2mx + m 2 2 = 0 a/ Xỏc nh m phng trỡnh cú hai nghim b/ Gi A= l nghim ca phng trỡnh, tỡm giỏ tr ln nht ca biu thc : 2 x1 x2 + x1 + x2 4 x1 , x2 10: I PHN CHUNG CHO TT C TH SINH ( 7 im ) Cõu I ( 1,0 ) : Tỡm tp xỏc nh ca hm s: a) y= x2 3 4x b) y= x +1 x 2x + 5 2 Cõu II ( 3,0 ) 1/ Cho parabol (P): y = ax2 + bx + c ( a 0 ) a/Tỡm a, b, c bit... ABC cú AB = AC = a v uuur uuu r uuu r uuu r uuur uuu r biu thc: T = AB.CB + CB.CA + AC.BA theo a ã BAC = 1200 Tớnh giỏ tr ca TRNG THPT THI PHIấN KIM TRA HC K I NM HC 2011 2012 MễN THI: TON - KHI LP 10 Thi gian: 90 phỳt (khụng k thi gian giao ) CHNH THC I/ PHN BT BUC (7,0 im) Cõu I (1,0 im): Cho cỏc tp hp: A = ( 5; 6 ) ; A B ; A\ B ; B \ A Cõu II (1,0 im): Gii phng trỡnh sau: Cõu III (2,0 im): Cho... ca biu thc: 7: I PHN CHUNG CHO TT C TH SINH ( 7 im ) Cõu I ( 2,0 ) 1/ Cho A = (0;2] v B = [1;4) Tỡm CR(A B) v CR(A B) 2/ Xỏc nh cỏc tp A v B bit rng : A B = {3,6,9} ; A\B = {1,5,7,8} ; B\A = {2 ,10} Cõu II ( 2,0 ) Cho hm s y = x2 + mx -3 (1) a/ Tỡm m th hm s (1) ct trc Ox ti im cú honh bng 3 b/ Lp bng bin thiờn v v th hm s (P) ca hm s (1) khi m = -3 c/ Tỡm to giao im ca th (P) vi ng thng... ( m n ) x + my = n (vi m, n l tham s v m2 + n 2 0 ) Tỡm iu kin ca m v n 2 ng thng trờn ct nhau ti mt im thuc trc honh 2 2 TRNG THPT THI PHIấN KIM TRA HC K I NM HC 2012 2013 MễN THI: TON - KHI LP 10 Thi gian: 90 phỳt (khụng k thi gian giao ) CHNH THC I/ PHN BT BUC (7,0 im) Cõu I (1,0 im): Tỡm tp hp A v B bit A B = ( -;5 , A \ B = ( ; 1 , B \ A = 3;5 Cõu II (2,0 im): Cho Parabol ( Pm) : y = x... i xng Lp bng bin thiờn v v ( Pm) vi m va tỡm c 2/ Tỡm m ng thng d : y = - x - 2 ct ( Pm) ti hai im phõn bit A, B sao cho OA vuụng gúc vi OB Cõu III (1,0 im): Gii phng trỡnh sau: 2 x x 3 x 2 + 3x + 10 + = x 1 x + 1 x2 1 Cõu IV (3,0 im): 1/ Trong h trc ta Oxy, cho A(1;-1); B(5;-3); C(2;0) a/ Chng minh A, B, C l 3 nh ca mt tam giỏc Tam giỏc ABC l tam giỏc gỡ? b/ CH l ng cao h t C ca tam giỏc ABC,... giỏ tr nh nht Cõu VII.a (1,0 im): Tỡm tp xỏc nh ca cỏc hm s sau : a/ y = 2 x 14 3x 2 5 x 2 2x b/ y = x + 3 + 3 x B/ Chng trỡnh nõng cao: 2 x + y = 5 Cõu V.b (1,0 im): Cho h phng trỡnh sau: x + 2 y = 10m + 5 Vi giỏ tr no ca m thỡ tớch hai nghim x.y t giỏ tr ln nht 2 Cõu VI.b (1,0 im): Cho phng trỡnh ( m + 1) x 2 ( m 1) x + m = 0 Tỡm m phng trỡnh cú 2 nghim phõn bit x1; x2 tha: x1 = 3x2 2 Cõu VII.b