đề thi thử dh hay số 67

so
GrAo DUC VA
EAOTAO HA NQr
rnr-IoI.rc
TIIPT CHU vAfq AN
{*
ne rlrt
THU DAr Hec EOTTr NAvr zor
r
MOn
Toin -
fndi A
Thdi
gian
ldm
bdi:
180
phft,
kh0ng td ttrOi
gian giao
d0.
OA ttri
g6m
01 trang.
PHAN cIruNG
(z
iti6m)
B il
(2
drd@.Cho hdm s5
y

=
?4
x-l
i. Kh6o s6t vd vc dd thi
cria
hdm
s6 da cho.
.
2.
Gei
r ra
giao
di6m't;ft;o't;;;e*;;
cria r10 th!. rim
di6m
M tr€n d0 thi sao cho ti6p
tuy€n tgi M
wdng
g6c
vdi
IM.
Bii
u
(2
diam)
lx-alvl+3
=
0
l. Giai hQ
phuong

trinh:
]
-
"'
[!/lo&,< /iog,y=o'
2.
Giaiphuong
trinh,
sit
2*
n
t-9t
t*
=
tan
x
-
cot x .
cosx smx
Bii
III
(1
die@. Tim th6
tfch
kh6i
trdn xoay dugc tpo thanh
khi
quay quanh
t4rc Ox hinh
ph8ng

gidi
han
bOi dd thi
(C)
cta
hdm s5
y
=
*dt([*) vd ciic cfudrng thing
y
=
0,
x
=
1.
Bei
W
(
did@.
Cho hinh ch6p S.ABC c6 d6y ld
tam
gi6c
cAn
(B
=
e
-
o)-
C6c rludng
thing SA,

SB, SC tpo vdi
m[t
phfrng
ddy cdc
g6c
b].ng nhau. Ggi V, ua V, lAn luqt la
th0
tfch cria
hinh ch6p
vd th6 tich
hinh n6n ngo4i tii5p
hinh ch6p. Tinh
theo cr ,y
,6
$.
'v2
(.
lxr+x+log21=8Y'+2Y+1
Bni V
Q
diA@. Giei he
phuong
trinh:
]
Y
lu'-ro+1=o
(vdix>
o'
Y>
o)'

L"
"4
PHAN
fV CHQN
Q
itid@. Thi sinh )nt
"npo
mQt trong hai
phb):
Phdn A ho(e Phdn B.
Phin A:
Bii
VIa
Q
diiim)
l. Trong mflt
phEng
tga ilQ
Oxy,
cho tluong trdn
{C):
(x-1)'
+(y +2)2
=9
vd
dutrng thing
(d):3x-4y+m=0.Timmd6tr6n(d)c6duynhfltdiOmPsaochotrlPc6thekedugchaiti6p
tuy6n PA, PB vdi
dudng
trdn vd tam

gi6c
PAB la 6m
gi6c
vudng
(A,
B ld cdcti€p diem).
2.Chohai
ducrngth$ng d'* 1=
yr
1
=t
^l
vd
A:
x-2
=
y,
3
=t
4
trongkhdng
212123
gian
v6i
hQ to4
dQ Oxyz. Bi6t rang d vd
Acft nhau. HEy vi6t
phuong
trinh
mf,t

phing
(P)
chrla
A
sao
cho
g6c gita
dulng th6ng d vd m6t
phing
(P)
l6n nh6t.
Bii
YIIa
(I
di6m). Gi6i
phuong
trinh: 9x
+
2(x
-
2).3"
+
2x
-
5
=
0.
PhAn B:
Bni vlb
(2

die@
1. Trong
mlt
ph&ng
tqa
dQ Oxy, cho
dudng
trdn (C): (x
-
l)t +
(y
+ l)'?
-
25
,
diiSm
M(7;
3).
Vi6t
phucrng
trinh ttuhng th&ng
qua
M c6t
(C)
tai
hai
di6m
phAn
biQt A, B
sao

cho MA
=
3MB.
2.
Tt cdc cht sd 2, 3, 4, 7, 8,
g
c6 th6 lflp
tluqo.
bao nhi€u
si5
tu nhien 16 c6 s6u cht s6, trong
d6c6dungbachfr s6ZZ
BAi VIIb
(
diim).Gi6i
phuong
trinh
tr€n tflp ttsp
f5
phttc:
za - z3
+
6*
-
8z
-
16=
0.
HCt
Hq vdtOnth{

sinh:
,
S6 Oao Oanh:

sent to
www.laisac.page.tl
Hr-l5Nc nAN cnA,r vn nrdu
odu
udx roAN
-
rsdl a
nnl
u6r nuNc
DIEM
BEi I
2d
CAU L
(1.25
ttiim)
a)TXD:x*1
0.25
b) Sg bi6n
thi€n:
+
Nh6nh v6
cqc vd duirng
tiOm cAn:
Iim
y=2'
lim

y=
2:'Dd
thi c6 tigm
cQnngang
y
=),
x-)-00
x-)+co
lim
y
=-oo;
lim
y
-
.o:
Dd
thi c6 tiQm cAn drfurg
x
=
1
x-+I x-+l'
o,25
_1
+
Yt= :
^
"
'
(x
-l)'

+
Bing bidn thit
F-
lY'
<
0,Vx
* 1,
do d6
hhm sd nghich bidn
uOn
(-co;
in
-oo 1 *co
I_
I
vd
(1;
+oo)
0,25
2
fco
+co
)
c) Dd thi:
Dd
thicSt
ox r?i didm
A(
j,
ol,

c6tOy tai B(0,
1)
VE dfng,
dep
0,50
Chu2
(0,75
ilidm)
Ta c61(L;2). Gi6 srt M(xo,
yo),
ta c6 hQ sd
g6c
ctra IM lb k
=
Y
o -2
=
l
-
.
xo-1
(*o-1)t'
4,25
Tidp
tuydn tai
M
vu0ng
g6c
vdi
IM khi

vd chi khi k.y'(x6)
=
-1
-1
1
e; :r J=
-1<>
(xs
-1)a
=
1
(xo
-l)'
(xo
-1)
o,25
TU d6 ta duoc M(2: 3) hodc M(0:1)
a"25
Bni u
2d
cau l,
(1ili6m)
+
Didu kiQn x > 1,
y
) 1. L{p ludn
dua vd hg:
[*-0,
+3:0
[loga

x
=logzy
0.50
i*-ay+3=0
<+1
r
Ix=y"
0.25
+
GiAi ra
c6
nshiOm
(1:
1):
(9:
3).
o.25
Cilr2
(1diim)
+
Didu kion
x *nL
'2
4,25
a-
+
Giai
(2)
ta duo.c *=
*

*k! no4c
x
=
n
+kLn
-tJ
_0al
o,25
+
Kgt hqp
didu
kiQn ta dugc
x
=tt+kZn
0,25
B}i IU
1 I
+
LAp
luan ra
dugc
y
=
*1G(i
+ *) khdng 0m
tr6n tdp xdcdinh 11
[O;+*)
;
0,25
I

+
Ldp luan
ra duo. c thd r(ch ld v
=
n
Ix2
h(l + x3)dx
0
0,25
+
Dung cong thrlc
rich
ph&n
tungphdn dd bidn
ddi tfch
ph6n
I
=
I*'h(l+
x3)dx
:
0
l=4.m6**rlll
-'r4.Ai
d*
=tn2-ti
,.t
o'.
3
'10

d3l+xr
3
jl+xi
0,25
+rrnh
dusc
tichphan J=
J**
=
i[.'
*jd"
1-1rn
z.ydvrhdrichcdntimld
v=
tr(21n2*r)
.
33
3
=d !,nrt**rill
3 3
',10
0,25
Bni w
1d
+
Ggi H ld chAn
ducrng cao cta
ch6p,
chrlng minh duoc
H

ld tam duong trdn ngo?i
1!{p
tamgi6c
ABC.
0,25
+
Dat AB
=
AC
=
a; Tinh dugc
dign
tich tam
gi6c
ABC li, S,
=
1a'sin2a;
2
0,25
+
Tinh dugc
diQn tfch
hinh
trdn
ngoai tiOp tam
gidc
ABC ld S,
na'
=-
4sin2

s,
0,25
+
Y
_
SH.q/3
_
S,
_
v2 sH.s2
/3 s2
4sin3
acosa
o,25
Bni
v
1d
+
(1)
e x3
+x+ log,
x
-log,
y
=
8y' +2y+1
<+
x3 +
x + log,
x

=
(2y)'
+2y
+logr(2y) (q)
4,25
+
L4p lufln duo. c
hbm sd f(t)
=
t3
+ t + 1og2
t
ddng
bion tren
(0;+oo)
0,25
+
Do d6
(3)
<+ x=2y.
4,25
+
Thd x
=
2y
vio
(2)
vb
kdt hqp didu ki0n
ta duo.c nghi€m *

=
l,y
=
1.
2
o,25
Bii VIa
2d
cau I
(l
rtidm)
+
Dudrng trbn
d6
cho
c6 Am
I(L;1\, b6n
kinh
R
=
3
;
0,25
+
Tam
gi6c
PAB
luOn
cAn tai P,
do d6 ndu n6 wOng

thi wdng
t+i P; Khi dd trl
ei6c
PAIB
lb
hinh
vu0ns
canh 3 vh do
d6 PI
=
3Ji
0,25
t*D€ t$l t+i
duy
nhdt didm P
thi
khotrng
cdch
tt
I den
(d)
bing
3J2
4,25
lm+81
+ Trlc
g Jil J
=3J2;
GiAi ra ta
duoc m

=
*8
xlsJt
5
0,25
cku?
(1
tlicm)
r.Fgr{s"-q" c eis-e-*" qgr.q.r+s 4":s"A l*"-M0;1-;"0-
;
Ldt;:t5;tfi;rilaCa;cal"ifid"Klan
dqr ie
**
*-;
-*t***-
hinh chidu
cria A trcn
(P)
vd A.
Ta c6
y
AH
< AK suy
ra
lAlvftI<
IAMK, ding
,
','l
,
=

,
thrlc
xAy
ra
khi vi chi
khi
H trung
K,
trlc
li
-
/
-r-
/ I
Q/
AKI(P).
/
'
/
lr
/
Nhu vay
g6c gifia
d vd
@)
lon
nh6t
khi
/
t

I
L1
/
vhchikhiAKJ-(P)t4iK.
/ffi*r\
0,25
4,25
+
Tim
duo.c toa dQ K
(h
hinh chieu
ciia A tr€n
A)
h
K
=(*ry-,?\
.rvrrA)LqLr
-[Zr
7t
7
)
0,25
+
Mar
phing
(p)
di
qua
K

vd
nhan
AR
=
f-:t:,:l
ldm
vrPr,
do
d6
phuong
\.
7'7'7
)
trinh cria
@)
li
9x
-3y
-
z
-
5
=
0
4,25
Bii YIIa
kl
+
DAt
t

=
3"
(t
> 0),
phuong
trinh fiA
thenh
tz + 2(x
*2)t
+2x
-
5
=
0
0,25
+
GiAiphuong
trinh dnttadud.c
t=
-l;
t
=
5
-2x
(1o4i
t
=
-1)
0,25
+Vdi

t
=
5-2x
tadugc
3*
=5*2x
(2)
X6t tfnh don diOu hai vd suv ra
(2\
cd khOne
qu6
m6t nshiOm
a,25
+ Mft kh6c x
=
1
thoi
mdn. VOy
phuong
trinh
de
cho c6 1
nghi€m
x
=-1-:-
0,25
Bii vlb
2d
Ceu
1

(1
rliim)
+
Dudng
trbn dd
cho c6 am I(1;-1),
b6n
kinh R
=
5;
MI
=
Jn
,5, do d6 M
nf,m neodi dulne trbn.
Q,25
;Ceil
Ie kh'A"s;a;,l';ti
dsnA;dns;hAns
MCil
irittruns
cidm
An,
ta co
I,A
:
I'B
=
fr5-;,
I,M

=
J52
-7
o,25
+
Tt MA
=
3MB
ta suy ra I'M
=
2I'A,
do
d6
Giiiraduo.cx=4.
0,25
;
Nhtvat dnhnt ih&ng
cdn tim
qua
M
vb c6ch I-mQt khoAng
bing
4.
TU
d6
tim duoc hai duirne
thoi
m6.n:
y
=

3; LZx
-
5y
-
69
=
0.
a.25
Chu2
(l
tliim)
Gi6
st sei tu
nhicn 16
d5 cho ld abcdef
,
+ C6 3 cdch chqn
f;
o,25
+
M6i
c6ch chgn f c6
Cl c6ch
chon
vi tri
cho
ba cht
sd 2;
a,25
+ M6i

c6ch
chqn f vh
chgn vi trf
cho
ba chfi sd
2 c6 5x5=25
c6ch
cho:t hai
cht sd
cbn lai:
0,25
+
Theo
Quy
t6c nhEn,
cd 3xC3rx25=750
sd
thoi mdn.
j
a,25
BEi VIIb
1d
+
(NhAn
bidt duo.c hai
nghi€m
z
=
-1,
z

=
2) Phuong
trinh
di cho
tudng
duong
v6r
(z-2)(z+I){2'?+
8)
=
0
+ Gi6i ra duoc
4
nghiom:
z
=
-10
z
=
2, z
=
XJS|
0,50
0;30

Hdt