Sat hạch 10-A lan 1-2012-Le Xoay-

\ Truong THPT L€ Xoay Nam hoc: 2Ol'1, - 2012 DE THI KHAO SAT CHUYEN DE LOP lO I,AN I Mdn:Tor{n-KhdiA+AB j' Thdi gian lim bii: 150 phrit (kh6ng kd thdi gian giao dd) ^ -t2 CAu I (2 ttidm): Cho Parabol (P) c6 phuong trinh: y = - '-vd didm MQ,-l). Gqi d li duong thing qua M '4 vi c6 hQ sd g6c li k. a. Chfng minh rlng vdi moi gLd tri crla k dudng rhing d lu6n cit (P) t+i hai didrn A, B phan biQt. b. Xdc dinh f*atictra k dd do4n AB ng6n nhflr. Cau II (3.5 tlidm): 1. Giii phuong trinh sau: xt +rl x' +I I =3I. .t (z 2. Giai he phuong trinh sau: lf^+ l'1+ 3x - 6v = 0 lx'+xy-3 3. Tim m dd phtrong trinh sau c6 3 nghiOm phan bi€t duong: x3 +2(m-x)(mx-l)=mx' CAu III (2.5 tlidm): Cho tam gi6c ABC bidt A(-2,3); B(4,3; vi C(1,0); Gqi G lii trgng tam cfra MBC vd I ld didm trOn canh AB sao cho L{=2IB a. Chrrng minh r[ng: MBC lb tam g!6c vuOng. ] b. X6c dinh tga dQ ctra didin D tren dudng thioe BC sao cho dulng thing ID song song vdi dudng thing AG. Cau IV (2 diim): r r sln2c cos'a ' 1. Unung intnn rang: I -:- = Sln d.cos a 1+cota I+tana (Vdi gii thi6t'cdc bidu thrlc dI cho ddu c6 nghla) 2. Cho a, b, c ld c6c sdthuc ducrng th6a mdn: a+b+c =J& Clrring nrinh rang: ab +bc + caZg(a + b + c) zags -___-__ I{o v} fdn thi sinh: . Sd bdo danh:. ( Cdn bQ coi thi kh6ng gitii thich gi th€m) oAp AN * THANG udvr Cdu v NOi dung Didm I a. 2 lho Parabol (P) c6 phuong trinh: y =+vi ttidm M(0,-t). Goi a li ttuong thing qua M rh c6 h€ so g6c li k. IMR: vdi moi gi6 tri cfia k tludng thing d ludn cit (P) tai hai diifn A, B phan bi6t. 1.0 - Phuong trinh dudng thang d li: y = lu-|. - Phuong tdnh hodnh d0 giao didm cira d vi (P) li: 2 -+ =/tx- 1 e x2 + 4lu- 4= 0 (l) 4 C6: A'=4k2 +4>0,VftelR. Do d6 phuong trinh (1) luOn c6 hai nghiQm phAn bi€t x,,xrv6i Vft e IR -+Duong thing d lu6n c6t (P) tai hai didm A, B phdn biOr v6i V& e IR 0.25 a.2s 0.25 0.25 b. Xdc ttinh gi6 tri cfia k dddogn AB ngin nhdt. 1.0 - Gii srl A(x,y,);B(x,yr) trong d6 x,,x, li hai nghi0rn ph0n biOt ctra ptr (1) t = .r;l"r-I; /" = fuz-l - Ap dgng dinh lf Vi- et ta c6: x,* x, = 4k;x,x, = -4 - Ta c6: AB2,= (*, - *r;' + (r, - yr)' = (t + tc,)(x, - ,r), = (t + k )[(", . *,,,)' - 4*,*,] =(t+k,)(an,+ro)>ro -+ AB24 DAu"-"xiy ra<+k=0 Vay vdi k=O,thi AB*,n = 4. 0.2s 0.25 0.25 0.25 il I rt+ ,f,'+1i=31. Giii phuong trinh sau 1.5 Dk: VxeR D+t Jr'z+f f = t, dk: r > 0 -+ xz =tz 11 Khi d6 phuong rinh:d6 cho tr6 rhlnh: tz + t - 42 = 0* [; = U' ,1,, , ^ v6it=6tac6: J7.lI =6+> x2 =25*[t=t. [x=-5 Vdy phuong trinh dd cho c6 hai nghi€m lh: x=5; x= -5 0.25 0.5 0.5 0.25 Giei h0 phuong trinh sau; Ta c6: (r) <+ {ff*+y)+ 3x-6y= o(l) \ / lx(x+ Y)=3(2) - Ta thdy x = 0 kh6ng ld nghi€m cira hQ. -XEt x* 0. Khi a6: (z) e x+ y =1 (3) The (3) vdo (i) ra dugc prr: U- *Z* - 6y = 0 <+ 3y, +3xz - 6ry = 0 ,-,8) o(t- !)'=0<)x=! Thay x - y vdo (2) ta duoc: 2y' =3 *> y' 3 E =t* Y:llt r.r=* V4y hq phuorng trinh d6 cho c6 hai nghiQrn li: (x,y) = [,E,1EJ,[ E ,rlz I; IJ - t_ ,rlz o.25 0.25 0.2s 4.25 m tli phuong trinh sau c6 3 nghi€m phfln biQt d Ta c6: (t) * x?(x-m)-2(x-m)(mx-t) = 0 T- _ <t (r- *)(*' -2mx+2)=o <+ l^;': lx'-2mx+2=A Q) Vdy phuong trinh (1) c6 ba nghidm phan biQt duong *, m > 0 vd phuong rrinh (2) c6 hai nghi€m phdn biOt duong kh6c m l*> o lo'to l. e{sto .'> I l"'o l*'-zm2 +2*0 Vay gi6 tri cdn tim cira mld: m, JZ . 0.25 0.25 Cho t1m gi6c ABC bidt ,4(-2,3); didm trdn canh AB sao cho IA = ZIB Ta c6: trB = (6,0), AC = (3, -3), gg = 1-3,-3) Khi d6: AC.BC = -9+9 = 0 *+ dC t gC . YQy MBC vuOng rai C. XD tga dQ cria diim D t.Cn f *^ _rn**u**, _=2+4+l _, l^c . ^ 1 ' r -+G(1,2) L, -le*la*lc =3*3*o =2 fto 3 3 - Vi t ttrugc cqnhAB mi IA = 2IB n€n ra c6: Vi lZW ( ^/ .l*, -xn=2(xu-x,) t 1 ly,-yn=2(yu-y,) - Goi toa dQ ctra didm D h: D(x,y) - Vi D tlruqc dqdng thing BC non 3_D ctrng phucmg vot Ee Mn BD=(x- 4;y-3); Ee =?3;-3) -+ x-i -+*"-y=l(l) '-3-3 - Theo gt h c6: ,ID cDng phuong vdi AG Md,,lD= (.r - 2; y - 3), 7G =(3; -l) - + = + -+, +3,y= I I (2) - Ta c6 tga dQ trgng tdfn G ctia tam gi6c ABC lh: Tu (1) vn (2) ta;c6 h€: 0.25 0.25 0.25 0.25 0.25 0.25 Chrrng minh rnng: , sin2 4 cos'a = Slll 4:COS 4 l+cota l+tana .1 1 . sln- 4 eos-a laco: vt -t _+ _r, sinl a + cos3a_, _ (sina+cosq)(sint a +eosta-sina.cosa) sina+cosa sina+cosa = t-(l-sina.cosa) = sina.cos a=VP (dpcm) Cho a, b, c Ii €6c sd thr;c duqng th6a m6n; a +b + c = Jobt Chung minh ringz ob +bc + ca) 9(a+b + c) Vdi moi sd thuc x, y, zta lu0n c6: x2 + y'.+ z' 2 xy + yz + zx +(x+y+z)' >3(xy+y+a) . Ap dqng BDT tr0n v6i x = ab; y = bc; z =ea ta duo. c: " (ob+be +c.a)i 23(,abzc+bcz a,t,+ce2b1=3abc:(CI+b+c) i -+ (ab + b c + ca) > Ji.J "b".J a + b + c = Jj.@ + b + clJi a a 111 :] rl l . Ir , 13 Mrt kh6c ta c6: a +b + c = mZ' lk,jjj,jf_ = (a +b * dW_ -+ Vc+b+p >3J3(2) Tt (1) ve (2) ta c6: ,ab + bc+ ca > g(a + b + c) (dpcm) Dau "-r' f6yrakhi vichilthi:,a= g=s-= !. :'' i' 0.25 0.25 ( Thf qinh lim theo c6eh khdc ddng v6n cho didm rfii da) tdtr g Ra ttd virlipr66- 5o 1 L= Nguydnfhi Hdng NgwydaThlEqnh . Truong THPT L€ Xoay Nam hoc: 2Ol'1, - 2012 DE THI KHAO SAT CHUYEN DE LOP lO I,AN I Mdn:Tor{n-KhdiA+AB j' Thdi gian lim bii: 150 phrit (kh6ng