Đề kiểm tra học kỳ 1 Toán 9 năm 2017 – 2018 phòng GD&ĐT Vĩnh Yên – Vĩnh Phúc

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PHdNG GD&DT viNH ynx or cninrn rHrIc Br rmnn rRA Heq rcp I NAnn Hgc z0t7 _2018 prON: roAN p Thdi gian: 90 phfit (kh6ng *A tnai gian giao dd) Cau 2. voi gi6 rri ndo cua athi hdm s6 y = (j, _.jj), _ J3 nghlch bi6n trdn R: a.. =f *. ".f c. ur S o. u. @ c6u 3. cho tam gi6c ABC *ong t?r-A. Kihd"e dinh ndo sau d6y Lir sai? A. cotB : tanC B. sin2B + cos2C: 1 C. tanB = cotC D. sinB : cosC Cf,u 4' Cho du'cmg tdn (o; 10cm), ddy AB : l2crn.Kho*rg c6ch tu tem o dth d6y aB bftrg: A. 8cm B" 7cm C . 6crn D . 5cm rr. TU,LUaN (8 triSm) CAu 5 (2,0 itihm). a)Tfnh: a:r(u6-2;, r$-.2J.; b) Timxbi6t: 2+nEi.l ,=*. cfiu 6 {1,5 diam). cho biriu thfc u =( lfi:;:J; -2-*j, G -, \ xVx-l l--1*) x-I a) Rirt gon bi6u thfc A; b) Tim x d6 A < 0. C6u 7 (1,0 iliAm).. Cho hdm s6 bgcnhdty: (m+I)x - 2m+3 (t) a) Tim * dq hAm si5 ( I ) di qua di6m A'( t ; i); b) Tim m o6 ao thi hdm s5 1t; song song "6i oo thi hdm sO-y : 3x + I. .9fl'j-fi:0 liary. cho nta du<rng trd n (o; ry co ducmg kinh AB.Durrs ddy Ac: R vit nep tuyeli Bx vsi naa ducvng tron. Tia phdn gi6c cira g6c BAC "8t ociui u, cdrttia Bx tai P vd chtnhadulng trdnlAm O tai e. a) Chung minh ring BF : pA . pg; b) chrmg minh ring 4 di6m B, p, M, c crmg thuQc m6t ducrng trdn. c) Chung minh ring AB ta tiilp tuy6n.A" [.,f o] t'2 ) d) Euong thing AC cattia Bx tqi K. chrmg mi,h rFng Kp : 2.8p. cau 9 (0,5 efiam). cho ciic s6 ducrng a,b,c th6a mdn a + b + c= 3 .chfmg minh rang: abbcca3 r/.'+3 tfa2+3 Jbr+l-2 r. TRAC NGHTEM (2 tri6m) Ghi vdo bdi ldm chi mQt chfr cdi A, B, c hofic D tru6c cfru trd rdi iffing. Ciu 1. Girl tri c,ia --.,j-- bEns: 47-1 e t" -lt '! r \. v / -t 8.1-"J7 c.-J7 -r a.,ll +t ..H6t (Gido vihn coi ki*m *a kh6ng gidi thich gi th*m)PHoNG Gp&pr viNH ytN ruAN rnA Hr.IoNG oAN cnApr pp rrnvr rRA Hec ry r NAnn ugc t 2ot7 - zotr tvt6u: roAN q CNG : TSng 2 di,5rn, m6i ceu drin o-s at6r" Cdu I 2 a J 4 Diry 6n D C B A II. rflAN r U LUAN:(8,O cli NOi du a) A: JGqf +Ji-2Jl = z - Jl* {.6 - 1; 2 = 2 -..6 *.,6 - I = 1 2+J2x-1-.r Jzx-t-x_z (t) [", I DK: J""-2ex>2 (2) l*>2 Phuong trinh (l) e Jrx -L = x - 2 e2x-l :(x -2)' e x' -6x+5=0 e (x-l).(x-5) =0o ['=, [x=5 f6t hqp vli (2) ra duoc x: 5 0,25 0,75 _l J;(,+Ji+r) 2-J;l ,_r -L@"G{jG_; -( J; ,2-.E.1 (J;+r)(.[-r) =t-T-t.- -[Ji-r' J;-)-Gj _ _J_ (.'6. rXG - r) _ z(G + r) -tr-l Jx-2 Ji-z 0,75 0,25 0.25 0,25 Hdm sd (t) di qua rti6m A(M 2 :(*+ I).1 - 2m* 3 o m: 2. yQy m:2 Ciu Y Di6n 5 a 1,0 b 6 l' b A<o*1i8., *l#., L* = O, x*t;x*4 |.* = O,x *l; x*4 fJ*- 2 <o <)j <+0<x<4:x*1 [x>tl;x+l;x*4 Vayvdi 0<x<4 thiA<0. 7 a 0,5