\Prr oNG-c D &Dr-l{INu YtN oE xrru rRA HQC rci' u NAU Hqc 2017 - 2a18
nt0N: roAN g
Thdi gian ldnt bhi 90 phtit, khhng kA thdi gian giao di'
I. TRAC NGHIpM (2,0 tli6m)
Ghi vdo bdi tdm chi mQt cttib ctii A, B, C hofic D tlwdc yhuong tin trd ldi drtng.
cAu 1. Phuong trinh nio sau t16y li phucrng trinh bac ntr6t rrai ry *,y?
A.2x+5y2=1g B.Zxy+Sy=lg C.2+5=10 D.2x+5y=10
Ciu Z.Cho parabo-l (f) ,y =+vi dudng thang(d) :! =Zx-3cit nhau tai hai di6m A vi B'
Khi d6, diQn tfch tam giitc OAB b{ng:
A.6 (dvdt) 8.7 (dvdt) c. s (dvd| ' D.9 (dvdt)
cAu 3. Cho rtudmg trdn (o;n) c6 AB =JZn DO dai cung nh6 AB bing:
c"ry
BE CHINH THUC
A. nR
c6u 4. Cho mQt hinh cAu c6 thc tich bang 50:o ,rr'. Di€n tich mflt ciu d6 1i:
3
5002 a
["
-
gnlz
3
u. PrrAN Tu LU4.N'(8,0 cli6m) (mx-v=l - ..1
ciu 5 (1,5 tri6m). cho hQ phuong trinh: IT^;;, : z (*,v ln 6n; lrr li tham sd)
a) Gi6i hQ Phuong trinh vdi m = -3;
b) Tim m d0 hQ cl nghiQm duy nhat (r;y) sao cho x vit y deu nh4n gi6 tri UY*n
Ciu 6 (1,5 tliam)Cho phuong trinh: x' -Zna+4m-4= 0 (1), (x li dn; mli tham sO)
a) Tim m d6 phuong trinh (1) c6 hai nghiQm ph6n biQt;
b) Tim m d6 hai nghiQm x,;x, cria phucrng trinh (1) th6a rndn:
( n-2)(*'-2)=x" +x" -8'
CAu 7 (1,5 rli6m). MOt ca nd xudi ddng tr6n mOt t<tric sing 1) biin A d6n b6n B dai 80 km, sau
d6 lai ngogr ddng rtiin dia di€m C c6ch btin B 721<rrL. f-trij Si-an ?u i9 xu6i ddng ft hon thdi
;i"r rdnr aOng Ia tS pftit. Tinh v6n t5c ri6ng cria ca nO bi6t vfn t6c ddng nudc le 4 km/h.
cAu 8
3;l t[i* .odn
(o)ngo4i tirip ram gi6c nhen ABC
.c6c du&ns cao AD, BE, cF cta
tam gi6c ABC cht nhau tPi I{.
a) Chrmg minh rAng tr: gi6c BDHF vd BCEF n6i tiiSp dudmg trtn;
b) Chring rninh rlngFH li tiaphfln gi6c cria g6c DFE;
c) Gqi MldtrungdiErn cua aC. Cfrung rninh rang tu gi6c DMEFnQi. ti6p dudrng trdn;
d) Gqi K h gial di6m cria dudrng thAng EF vitBC. Chung minh rang H Id tr.uc tdm cria
tam gihc AMK.
CAu 9 (0,5 di6m)
Cho ba sii ducrng a,b,c th6a mdn a + b + " = JM. Tim gi6 tri l6n nh6t cria bi6u thric
-=L- 2b - 2c
- Jo' +2018 r/b' +2018 Jc2 +2018
- - -- -- - - -- H
e t --- - -- --'
Gido vihn coi kidm ta khdng gidi thich gi thAm.
B.ZrR
8.50r cm2 C.25tt cltt2
nR
D.-
4
D. tA}r cm2
Ho vd ftn lqc sinh: SBD:
PHONG GD&ET \IINH YEN
I. TRic NGHIEM qz,o aiaml: ME
II. Tr/ LUaN 18,0 diOm1.
HDC KIEM TRA Hgg rci, rr xAm Hgc zafi - 2018
MON: roAx q
em
oB cnixH THUc
: M6i c6u drine 0.5 di
CAu I2J4
D6p 6n DACD
CAU NQi dung Di0m
5 (1,5
cli6m).
a) Vdi ln : -3 thi he phucrng trinh tro thinh
Ir
l-l*- ! =7- l'= -t I
12*-3y=2 1.._ 8
l.'--11
NghiQrn cria hQ phuong trinh li (x,y' ( t 8 \
)=[-11;-n)
0r75
fntx-y=1 fy=ntx-L fy=mx-L
l.2x+ nry-2 [2x+ nt(ntx-1)=2 '- [(m'+Z)x=m+2
I nt+2 I m+2
l V = tn._;_ _l l X =--___-__
l" m'+2 I mz +2
1 <=<
I rn+2 I 2m-Z
lX=- l1r---
I tn'+2 [' m2+2
b)
0,25
ddu duong
lm+2>0 lnt> -z
<>< <i{
l2nt-2>0 " l*rl
V6i rnoi gi6 tr! cria ra hQ
khi: I m+2
trro._.J*= r;*1>tt
lyro- I 2nt-2
\r' lY=:--:-:>Q
t- ln'+Z
pt c6 nghiQrn duy nhdt. Ta c6 x vit y
e m>1.
0r75
0r5
6 (1,5
eli6m).
a) A = (nt-2)2
Phucrng trinh ( 1) c6 hai nghiQrr phAn biQr x, ;x,
khi: A' > 0 e(m-2)' r0e m*2.
b) Vdi moi m thi phucrng trinh (1) lu6n c6 hai nghiern-
Theo h6 thric vi - dt, ta c6: {x, + x, = 2nt
|.', '', =4nt-4
(x, -2)(x, -Z)=x,'+xr' -8e xtxz-2(n*xr)+4=(x, + *r)'-2xrxr-g
o (r, **r)' -3x,xr+2(xr+xr)=12
0,25
(2m)' - 3.(4nt - 4) + 2.2m = t2 e 4m(nt - 2) = 0
fm=O
^I lnt-)
L"'
:0 hodc m:2.
Ta co:
vay m0r5