CNG ễN TP HC K I NM 2014 2015 TRNG THPT PHC TH MễN: TON LP 10 PHN I : AI S I NI DUNG ễN TP i s Mnh - hp: Cỏc phộp toỏn giao, hp, hiu ca hp Hm s bc 2: Tỡm h s a,b,c parabol y = ax + bx + c (hay vit phng trỡnh parabol) tha iu kin cho trc Xột s bin thiờn v v th hm s bc hai: Phng trỡnh h phng trỡnh: Gii phng trỡnh cha cn, cha giỏ tr tuyt i dng n gin Gii h phng trỡnh gm mt phng bc nht hai n v mt phng trỡnh bc hai hai n; h phng trỡnh i xng Gii v bin lun h hai phng trỡnh bc nht n Tỡm iu kin ca tham s h phng trỡnh i xng tha iu kin cho trc Bt ng thc: Chng minh bt ng thc Tỡm GTLN, GTNN So sỏnh giỏ tr cỏc biu thc II BI TP Mnh - Tp hp Bi 1: Cho trc s: A = [4;9], B = ( 0; + ) , C = ( ;5] a A B, A C b A \ B, B \ C c d Xỏc nh cỏc hp sau v biu din trờn A \ ( B C ); ( A \ B ) C Ă \ B; Ă \ ( A B ) Bi 2: Cho A = { x R : x 8} , B = { x R : x < 7} Xỏc nh cỏc A B, A \ B; AB Bi 3: Cho hai hp: A=[1; 4); B = { x R / x 3} Hóy xỏc nh cỏc hp: A B, A \ B, A B Hm s bc hai Bi 1: Cho hm s: f ( x ) = ax + bx + c cú th (P) a) Xỏc nh hm s bit (P) l mt parabol cú nh S(2; 1) v i qua im M(1; 0) b) Kho sỏt s bin thiờn v v th hm s va tỡm c Bi 2: Xỏc nh hm s bc hai : y = ax + bx 1, bit rng th ca nú l parabol cú trc i xng l ng thng x= v i qua im A(1; 6) Bi 3: Xỏc nh cỏc h s a,c ca parabol (P): y = ax x + c , bit (P) i qua im M ( 2;1) v cú honh nh l V parabol (P) vi a, c va tỡm c Bi 4: Cho hm s: y = ax2 + bx + a) Xỏc nh a, b bit th hm s l mt parabol cú nh I(3;4) b) Xột s bin thiờn v v th hm s va tỡm c Bi 5: Tỡm (P) : y = ax + bx + , bit (P) i qua A ( 1;6 ) v cú tung nh l Bi 6: Cho hm s y = ax2 + bx + a) Xỏc nh a, b ca hm s bit th hm s i qua A(1;0) v B(2;15) b) V th hm s va tỡm c cõu a Bi 7: Cho hm s y = x + 3x ( P) a) V (P) b) T th hm s bin lun s nghim ca pt x + 3x = m (*) c) tỡm m (*)cú nghim phõn bit nh hn Bi 8: Xỏc nh Parabol (P), bit (P): I(1; 4) y = ax + bx + c i qua im A(2; 3) v cú nh Phng trỡnh H phng trỡnh Bi 1: Gii cỏc phng trỡnh sau : a d 5x + = x x + x + 11 = x b e 3x = x 4x + x2 = x + c x + x = f x + 3x = x + g -x + x + + x = h x x + x x + = 10 k x = x + +1 l x = x2 2x i x2 + 2x2 x + = 2x n 2x + x + = Bi 2: Gii cỏc phng trỡnh sau: x = x2 2x a d x + = 5x + 6| g b c |x + 3| = 2x + 2x + = x e x = |3x2 x 2| x f |x2 2x| = |x2 x + x 15 = x + Bi 3: Gii h phng trỡnh: a) x + y = 2 x + y xy = b) x + y + xy = x + y + xy = xy + x + y = c) x + y + x + y = d) x + y = 2 x + y + xy = 13 Bi 4: Gii h phng trỡnh: a) x = y y = x x = 5x+y y = y +x b) Bi 5: Cho h phng trỡnh c) x = 3x+2y y = y +2x d) y2 + y = x2 x = x + y2 x y = x + y = m a) Gii h m =10 b) Gii v bin lun h phng trỡnh ó cho Bi 6: Cho h x y = 2 x + y = m a)Gii h vi m = 26; b)Tỡm m h cú nghim nht Bi 7: Cho h x + y = 2 x + y + mxy = 13 a) Tỡm m h phng trỡnh vụ nghim b) Tỡm m h cú nghim nht Bi 8: Gii v bin lun h phng trỡnh: a mx + y = m + x + my = b x + my = mx 3my = 2m + Bt ng thc Bi 1: Chng minh rng: a, b, c c (m 1) x + (m + 1) y = m (3 m) x + y = a) ( a+b ) ( a + b2 ) b) ( a+b+c ) ( a2 + b2 + c2 ) Bi 2: Chng minh rng: x + y x y + xy , x > 0, y > Bi 3: Cho a, b, c l di cnh ca mt tam giỏc Chng minh rng: a + b + c < 2(ab + bc + ac) Bi 4: Cho a, b, c >0 Chng minh rng: Bi 5: So sỏnh A v B, bit Bi 6: Cho a + b3 = a+b b+c a+c + + c a b A = 2015 2014, B = 2014 2013 Chng minh rng: a + b Bi 7: Tỡm GTNN ca hm s: a) f ( x) = x 2006 + x 2007 Bi 8: Tỡm GTLN ca Bi 9: Cho a) a, b, c > b) f ( x) = x + 2x + c) f ( x) = x x + 2014 x2 f ( x ) = 3x + 12 x Chng minh rng : a b c + + ; b+c c+a a+b b) a2 b2 c2 a+b+c + + b+c c+a a+b PHN II: HèNH HC I NI DUNG ễN TP Cỏc nh ngha Vect l mt on thng cú hng Kớ hiu vect cú im u A, im cui uuu r B l AB Giỏ ca vect l ng thng cha vect ú udi ca vect l khong cỏch gia im u v im cui ca vect, kớ uu r hiu AB r Vect khụng l vect cú im u v im cui trựng nhau, kớ hiu Hai vect gl cựng phng nu giỏ ca chỳng song song hoc trựng Hai vect cựng phng cú th cựng hng hoc ngc hng Hai vect gl bng nu chỳng cựng hng v cú cựng di Chỳ ý: + Ta cũn s dng kớ hiu r r a , b , biu din vect + Qui c: Vect r cựng phng, cựng hng vi mi vect + iu kin cn v im phõn bit A, B, C thng hng l hai vộct uuu r uuur AB , AC cựng phng Cỏc phộp toỏn trờn vect a) Tng ca hai vect uuu r uuu r uuur AB + BC = AC uuu r uuur uuur ABCD l hỡnh bỡnh hnh, ta cú: AB + AD = AC r r r r r ( ar + b ) + cr = ar + ( b + cr ) ; a+0=a Qui tc ba im: Vi ba im A, B, C tu ý, ta cú: Qui tc hỡnh bỡnh hnh: Vi r r r r a+b =b+a; Tớnh cht: b) Hiu ca hai vect Vect i ca Vect i ca r r r r a b = a + ( b ) r a r l vect r b cho r r r a +b = Kớ hiu vect i ca r a l r a r l Qui tc ba im: Vi ba im O, A, B tu ý, ta cú: uuu r uuu r uuu r OB OA = AB c) Tớch ca mt vect vi mt s Cho vect r a + r ka + r r ka = k a v s k R cựng hng vi Tớnh cht: iu kin iu kin r a r ka l mt vect c xỏc nh nh sau: nu k 0, r ka ngc hng vi r a nu k < r r r r r r r r r k ( la ) = (kl)a k ( a + b ) = ka + kb ; (k + l)a = ka + la ; r r r r ka = k = hoc a = r r r r r r hai vect cựng phng: a vaứ b ( a ) cuứng phửụng k R : b = ka uuu r uuur ba im thng hng: A, B, C thng hng k 0: AB = k AC Biu th mt rvect theo hai vect khụng cựng phng: Cho hai vect khụng r r r cựng phng a , b v x tu ý Khi ú ! m, n R: xr = mar + nb Chỳ ý: H thc trung im on thng: M l trung im ca on thng AB uuur uuur r MA + MB = H thc trng tõm tam giỏc: G l trng tõm ABC Trc to uuu r uuu r uuur r GA + GB + GC = uuu r uuu r uuur OA + OB = 2OM uuu r uuu r uuur uuur OA + OB + OC = 3OG (O tu ý) (O tu ý) Trc to (trc) l mt ng thng trờn ú ó xỏc nh mt im gc O v r r mt vect n v e Kớ hiu ( O; e ) r r r u = (a) u = a.e uuur r M (k ) OM = k e uuu r r trc: AB = a AB = a.e To ca vect trờn trc: To ca im trờn trc: di i s ca vect trờn Chỳ ý: + Nu cựng hng vi thỡ Nu ngc hng vi thỡ + Nu A(a), B(b) thỡ AB = AB AB = AB AB = b a + H thc Sal: Vi A, B, C tu ý trờn trc, ta cú: AB + BC = AC H trc to H gm hai trc to Ox, Oy vuụng gúc vi Vect n v trờn Ox, Oy ln r r lt l i , j O l gc to , Ox l trc honh, Oy l trc tung To ca vect i vi h trc to : To ca im i vi h trc to : Tớnh cht: Cho +) r r a = ( x; y ), b = ( x ; y ), k R , A( x A ; y A ), B( xB ; yB ), C ( xC ; yC ) : x = x r r a=b y = y r + ) b cựng phng vi 0) +) r r r r u = ( x; y ) u = x.i + y j uuur r r M ( x; y ) OM = x.i + y j r +) ar b = ( x x; y y ) r r a0 k R: +) r ka = (kx; ky ) x = kx vaứ y = ky x y = x y (nu x 0, y uuu r AB = ( xB x A ; yB y A ) +) To trung im I ca on thng AB: xI = + )To trng tõm G ca tam giỏc ABC: xG = x A + xB y + yB ; yI = A 2 x A + x B + xC +) To im M chia on AB theo t s k 1: ( M chia on AB theo t s k xM = uuur uuur MA = k MB ) ; yG = y A + yB + yC x A kxB y kyB ; yM = A k k II BI TP Bi 1: Cho t giỏc ABCD Gi I, J ln lt l trung im AC v BD Gi E l trung im I J CMR: uuu r uuu r uuur uuur r EA + EB + EC + ED = Bi 2: Cho tam giỏc ABC vi M, N, P l trung im AB, BC, CA CMR: uuur uuu r uuuu r r =0; uuur a) AN + BP + CM uuuu r uuur b) AN = AM + AP ; c) uuuu r uuur uuu r r AM + BN + CP = Bi : Cho hỡnh bỡnh hnh ABCD tõm O Gi I v J l trung im ca BC, CD a, CMR uur uuur uuur AI = AD + AB ( ) b, CMR uuu r uur uur r OA + OI + OJ = Bi 4: Cho tam giỏc MNP cú MQ l trung tuyn ca tam giỏc.Gi R l trung im ca MQ Cmr : uuur uuu r uur r a ) RM + RN + RP = uuur uuur uuu r uuu r b) ON + 2OM + OP = 4OR , O c) Dng im S cho t giỏc MNPS l hỡnh bỡnh hnh Chng t rng uuu r uuur uuur uuur MS + MN PM = MP d)Vi im O tựy ý, hóy chng minh rng uuur uuuu r uuur uuu r uur uuur uuu r uuuu r uuu r ON + OS = OM + OP ; ON + OM + OP + OS = 4OI Bi 5: Cho tam giỏc MNP cú MQ, NS, PI ln lt l trung tuyn ca tam giỏc a) Chng minh rng: uuur uuu r uur r MQ + NS + PI = b) Chng minh rng hai tam giỏc MNP v tam giỏc SQI cú cựng trng tõm c) Gi M l im i xng vi M qua N , Nl im i xng vi N qua P , P l im i xng vi P qua M Chng minh rng vi mi im O bt kỡ ta luụn cú: r uuuur uuur uuur uuur uuu r uuuu ON + OM + OP = ON ' + OM ' + OP ' Bi 6*: 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 a ) CMR: AK= AB + AC b) Gi D l trung im ca BC Cmr uuur uuuu r uuur KD= AB + AC Bi 7*: a) Cho MK v NQ l trungrtuyn ca tam giỏc MNP Hóy phõn tớch cỏc vộct uuur uuur uuu r uuur r uuuu r MN , NP, PM theo hai vộct u = MK , v = NQ uuu r uur b) Trờn ng thng NP ca tam giỏc MNP ly mt im S cho SN = 3SP Hóy phõn uuur r uuuu r r uuur tớch vộct MS theo hai vộct u = MN , v = MP Bi 8: Cho tam giỏc ABC a, Hóy dng cỏc im P, Q cho uuur uuur AP, AQ b, Biu din theo uuur uuuu r AB, AC uuu r uuu r uuu r uuur r PA = PB, QA + 2QC = c, Chng minh PQ i qua trng tõm tam giỏc ABC Bi 9: Cho : uuu r r r uuu r r r uuur r r OA = i j , OB = 5i j , OC = 3i + j a) Tỡm ta trng tõm, trung im cnh AC ca tam giỏc ABC b) Tỡm to ca cỏc vect r c) Xột a = (2; y ) Tỡm y ngc hng r uuur uuur uuur AB v u = AB 3BC r uuur a cựng phng vi AB Khi ú r a v uuur AB cựng hng hay Bi 10: Cho im A ( 3; 1) , B ( 2; ) , C ( 5;3) a) Tỡm D cho t giỏc ABCD l hỡnh bỡnh hnh b) Tỡm M cho C l trng tõm tam giỏc ABM c) Tỡm N cho tam giỏc ABN vuụng cõn ti N d) Tớnh gúc B Bi 11: Cho im A ( 1; 1) , B ( 1; ) , C ( 3; ) a) Cmr ba im A, B, C lp thnh mt tam giỏc b) Tớnh di cnh ca tam giỏc ABC c) CM ABC vuụng Tớnh chu vi v din tớch ABC d) Tớnh AB AC v cos A e) Tỡm ta chõn ng phõn giỏc ca gúc A ca tam giỏc ABC Bi 12: Cho A(2:3), B(1;1), C(3;3) a) CMR tam giỏc ABC cõn.; b) Tớnh din tớch tam giỏc ABC c) Tỡm ta trc tõm tam giỏc ABC d) Tỡm ta hỡnh chiu ca A trờn BC e) Tỡm tõm ng trũn ngoi tip tam giỏc ABC Bi 13: Cho A(3;2), B(4;3) a) Tỡm M Ox cho tam giỏc MAB vuụng ti M b) Tớnh din tớch tam giỏc MAB c) Tỡm D cho t giỏc MABD l hỡnh bỡnh hnh d) Tỡm E (1; x) cho A, B, E thng hng