Đề kiểm tra học kì 1 (HK1) lớp 9 môn Toán năm 2017 2018 phòng GD ĐT Vĩnh Yên Vĩnh Phúc

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 aJ 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

b EC dO thi hdm sd (1) song song v6i dd thihdm s6 y: 3x+l thi: € m-Z.Y|y m:2. lm+1:3 {m:2 ) ^) I-r*+3*l-l*+t 0,5

8

a Ta co AAQB n6i ti6p ducmg tron ducrng kinh AB + AAQB vudng t4i Q = BQ.l-AP Ap dpng hC thfc giira canh vd ducrng cao vdo A ABP vudng tpi B dudng cao BQ ta co BP- : PA . PQ

0,5

0,5

b AC : AO: R + AACO cdn tai A md AM lA ph6n gi6c = AM ld rludng cao n6n 6M0= 900vd fdd =900 ( do Bx ld ti6p tuy6n cria dudng trdn)=B vd M cung thudc ducrng trdn dudng kfnh Op. Vpy 4 di6m B, P, M, O cirng thuQc mQt ducyng trd;

1,0

c Ke tlucrng cao CH vu6ng g6c v6i AB tai H.

rinh CH: + R = H .f ',*nl. rr,la ABrcH ( HeAB) 2 [ .2 _) . / l- \ v6y AB ld ti6p tuytin cira I c;$n l. 1..2)

0,5

d Ta c6 AAOC dAu 2 = 600

Xet A AKB vu6ng: cos I = !!- - .4K = AB-,.^ = 4R AK cos60u

AP lA ducrng phdn gi6c n6n { =++ PK =2BP BP AB

0,5

9 (a+b+c\' Ta co > ab+bc+ca+ ab+bc+ ca<3 aJ Ta c6: ab ab /_ tlr' +3 "lc' + ab + bc + ca c2 + ab+ bc +ca J(o*c)(a+c) -ab( I I \ j-t -r- | 2\a+c b+c)

vr<L( ob * ob * b' n'o * t' ')= 2\a+c b+c c+a c+b a+b) (tlpcm)

D6u bane x6y ra khi a: b: c : I

*@*t+c)=1

0,25

0,25

Chrt i,: - Hqc sinh gidi cdch khdc n€u dttng vdn cho dt€m tiii da.

- Hoc sinh kh6ngvd hinh hodc vd hinh sai thi kh6ng chiim bdi hinh.