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8/17/2019 Cac de Thi Tren Bao THTT
1/18
~ E
s6
(Thi?igian Zhm bhi: 18Ophtit)
a) KhHo sat SIJ b i b
h i h
vii vE?d , i, h h
~ O .
b) ~ i k thucmg
Mnh
d u h g thhg d song song
v6iduhgthhgx+y-2=~v~c~t C)tqihai
dibm A B p h b biet sao cho tam giac
L4B
c6
dien tich 2 , v i a giao km hai tiem c h .
Cau
2
1
d i h .
GiAi phucmg
trinh
(Xu 3 1
&).
Tinh gi hpn
L
=
lim
?I1+3x-1-xJ1-x
x O
x3
+
x2
c r u 4 16 .
a) Co d t cH bao nhieu
C I
vq ch ng hgc hien
viec bit tay l h nhau (dt nhih m8i nguhi
Caw 1
A .
rong khiing gian v i he .mc
t ~ aij -Oxyz, cho hinh ch6p tam giac d u
S M C v i A(3;O;O); B(0;3;0) vii C thuijc tia
Oz im t ~ aij dikm S bikt thk tich ci, kh6i
ch6p
S DC
b h g
9.
CPu
6
1 &). Cho hinh chop S.ABC c6 dhy
ABC lh t m gihc d u
t ;
hinh chiku cfia S
t r h mat day lh trung dikm ah doqn
thhg
AO.
~ i k t SO
=
a vA SAB la
tam
gihc
vuiing.
Tinh
theo a thb tich ci, kh6i ch6p
S.ABC
vh khohg c h h
t r
t d u h g trbn
n g o ~i tamgik SAC d b mbt phhg SCO.
h
1
h .
rong a t
~ g
ihe tr lc@a
dij Oxy, cho d u h g trbn
(0: y2
= 2x Tam
gik ABC vuiing tqi
A
c6AC 11
ti
tuY cfia
(C)
bong 86
A
la
ti
h,
b u h g cao
k6
t r A la H(2;O). T h @ad4j dinh B c h tam gih
2
ABC bikt
B
c6 tung dij ducmg va SABC=
-.
&
Caa 8 1 d i h ) . GiHi he phucmgMnh
8/17/2019 Cac de Thi Tren Bao THTT
2/18
sac TRUWt THI
mf s6
Thag i an ldm bbi: 180phcit)
C ~ U (2 di&). C ~ Osmd = 2 - d + + m
thhg
6 2
vikt phuung d u h g thhg A di
qua
(m la tham s6) c6 db thi
C,).
M ,
vu6ng g h
viri
dl
v i
tqo
vcii
r t phibg
a) KhAo sSt u b i h thih
va vb
d6 thj cCa him
P) mot g6c 30'.
s6
C)
hi
m
=
0.
Cau
6
1
didm).
Cho
in
ch6p
S.ABCD
c6
\
dky
ABCD la in vuhg, SA ABCD),
b)
~ i m tk
tbn tqi ti tuyh
viri ti3
SA = a . D i h tich
t m
giSc SBC b b g
(C,)
di
qua
d i h
A(3;O) va
cht d u h g tron
a2 Z
.
( S ) c6 phucmg trin ( x+1)2 ( y )2= 25
2
Tinh th6 tich kh6i ch6p
S.ABCD
theomot diiy cung M N c6 do dhi nh6 nhht.
Cau
1
didm).
GiAi phumg
trin
c o s 4 x - f i s i n 2 x + 2 = .
sin
4 x
os
2x
Clu
3
1
didm).Tinh tich p h h
Clu
4
1 didrn).
theo a. GQi I , J l luqt 1 i trung di mCSC
cpnh
SB v i SD.
Tinh khohng c k h
gifts
hai
d u h g hbg AI va
CJ.
Cau
7 ( 1 di&). Trong mat phhg v i he Qa
do
Oxy
cho
in
thoi ABCD a t n I (2; 1 )
thhg AB, N ( 0 ; 7) thuoc d u h g thhg CD.
Tim t ~ ao di6m
P
bi6t ng =
5 E va
diim B c6 tung do ducmg.
a) GiAi phucmg
trin
Clu 8
1
bibm).
GiAi he phucmgtrin
2
1
1
2
4) =log,,5 ( x + 2) +-1 2015(+-3)~
4 [ m + = - J y T ; ; = r .
8/17/2019 Cac de Thi Tren Bao THTT
3/18
8/17/2019 Cac de Thi Tren Bao THTT
4/18
8/17/2019 Cac de Thi Tren Bao THTT
5/18
sac TRU& Kt THI
olis6 s
( T h agi n
ldm b 18
phcit)
C U (2 d m ) .Cho h b y = x4 mx2+1 .
1)
~ h bhtvsl vt dd thj h b 6khi m 1 .
2 ) Tim m d&hhn s6 c6 ba cvc tri d&ng h&i ba
di&m qc tri c6a dd thj him s6
tqo
thanh mot
tam
gihc
c6 dien tich b h g 4 .
X
Clu2 1 dl ). ~ i n bichphiin I =
j
JT ~d*.
Clu 3
1
diim). Cho hai s6 ph3c z l , z2 th6a
m8n lzll
=2, 12;?1 =5, l
- I =4. HHy tlnh
11 +
z21.
Ciu 4 1 d i h ) .Trong khang
gian
vCri he Wc
t ~ a o Oxyz, cho hai di&n A(-2; 2; O),
B(-1;
1 ;
-1)
vsl
m t phhg
9
c6 phucmg
Mnh
2x
+
2y
+ 2
0. HHy viet phucmg
trinh
mat
phhng Q) chira AB, vu6ng g6c v6i (P) vsl vibt
phuongtrlnh@&(flc6~~ti&&v6i(~.
Clu
5
1 dikm).GiAi phucmg
trinh
3 J 3 ' - 5
J F Z 2J5d-25 X
E
R .
CPu
6
1
di im) .
Cho
in
ch6p
SABCD
c6 dhy
ABCD lb
hinh
w6ng cgnh a.Hinh chiku w6ng
g c c6a S
len
mat dhy
lsl
di&m thuoc AB sao
cho BI
241.
G6c gi0a @t bib (SCD) va m t
phhg dhy b h g 60 . Tinh th&tich kh6i ch6p
SABCD
vsl
khohg chch gi0a hai dutmg thhg
AD vsl SC the0
a.
Cilu (1 d i m ) .Trong mat tga do
Oxy,
cho tam
gihc
ABC
w6ng c h
Qi
A,
c~
BC
c6 phucmg
Mnhx +
3y 3 0
Chc di&m
I(3;
I ,
J(4; 3 )
l h
luqt n h ren c6c d u h g thbg AB vA AC. Tim
@ado chc dinhc
am
gihc ABC bikt
r h g
dinh
A c6 h o u do
nh6
h m 3.
Clru 8 1 d i m ) .GiAi h$ phucmg
trinh
Clu 9 1 diim). Cho x, y, z la ba s6 thvc ducmg
th6am3nln2 1
V A Z I
1 . ~ l m ~ i i t r i n h 6 n h h t
4 -2
3
I
chbieuthirc M = - +
l + y l x 3(1+xy) '
NGU* QWG THI
(GVTHPT
Bho
Ldc, Lbm ng)
8/17/2019 Cac de Thi Tren Bao THTT
6/18
~ ~ C T R U W ~ ~ T H
~ ~6
ThiXgian
l im bhi:
180phcit
C u 1 2 dih). Gpi (Cm)lad8 thi c i a h h
p h u m g Mnh mat p h h g a ) c i t inat c h
s6 y
=
x3 x +
rn m
h tham s6 thpc).
a)
Khao sat su bi&n hiCn vh v&d8 thi c i a
hhms6khi
m = 2 .
b)
Xhc dinh tham s6
m
d&qua di&md n iia
d8 thj (Cm
k6
duqc mijt d u h g t h h g d)
4
vcri
d6
thi(Cm)mijt h in h p h in g ( H ) v h d)
t i ip tqc c h h rCn hai trqc tga dij mijt t m gihc
T ) sao cho di n tich ciia H) vh
T )
bibg
nhau d&ub h g 2 (dvdt)
.
CAu
1
dih) . Giii p h u m g
trinh
tanrcot2x= 1+sinx) 4cos2x+4sinx-5).
Cau
1
d i h ) . Tinh tich p h h
R
h ( 4 t a n x )
I I dx
sin 2x.h
(2
tanx)
CAu 4
1
diim).
a) Trong t n r h g hqp khai tri&n he0 nhi thuc
Newton c i a bisu thuc (1 x 2 r ta c6 h( s6
chua x8 bibg 210. Tinh t8ng c k h( s6 cha
chc s6 h w g duqc khai tri&n r bi&u hirc tren
theo t n r h g hqp 86.
b) Cho chc s6 phuc z th6a m h Iz
-
11= Z
(S)
:
x +
2)' +
(y 1)2+
Z
1)' =
5
the0 thi&t
di n la mijt d u h g trbn ma c6 di(n tich S
=
n .
COlu
6
1 d i h ) .
Cho hinh ch6p t13 giac S.ABCD, day ABCD 1h
hinh
vu6ng c&
a
cqnh
hen
SA
I
ABCD)
vh SA =o .Qua A d n g mat p h ~ g
a )
u6ng
g6c viri SC sao cho a ) i t SC, SB, SD l h
luqt Qi G,M
N.
Tinh
theo a th& ich kh6i n6n H), bi t Ang .
d u h g t d n bay ciia
H)
ngoai t i ip ht gihc
AMGN
vh
itin
cua
H)
n b rCn dhy
ABCD ciia hinh ch6p S.ABCD.
CAu 7 1 di m).
Trong mat p h h g viri
h
@c tpa dij
Oxy,
hiiy
tinh di n tich
t m
giac ABC bi&t r ibg hai
d i h H 5 ;5 ), I (5 ;4) l h uqt lh trgc am h
t n d u h g trdn ngoqi tikp tam gihc ABC vh
x + y - 8
=
0
1h p h u m g
trin
d u h g thing
chua canh BC ciia
t m
giic.
COlu
8
(1 dih) .
C u
9
1 d i h) .
~ h oa s6 d u m g x,y,z th6a r n h O < x < y < z .
8/17/2019 Cac de Thi Tren Bao THTT
7/18
T irStrcmrn~m
ds67
(Tkdigian l h ai: 180 phzit)
x / l a
iu 2
dikm). Cho h8m s6
y =
diy
ABC
lii tam giic c h dinh C; d u h g t h b g
BC
x - 2
Qo viri m t p h h g
(ABBY )
mijt g6c
a) hb s i t vP vi5 dB
thi
H) c h 6. 60°
A B = A A 1 = a .
Ggi M N l h luc$ 1P trung
b) vikt phumg Mnh ti tUYknc h H) bikt &Ig
ti
&iimc b c&
BB1,CC ,BC.Tinh
th&
~ h i
tuykn song song v&i U u h g t h h g A : x y =0.
l b g try
ABC.A B '
vii khoAng c b h giita hai d u h g
CAm 2 1
diim). GiAi c b phuong trln
t h h g
AM NP
theo
a .
PU7 1 dikm). Trong @t phibg Oxy, cho
hinh
chit
nhgt
ABCD
c6 phuung trinh cgnh AB 1P
x - 3 y + 5
GO, phuung
trinh
d u h g
chb
BD
l n x - 4
l l
x - 1=
0;
bikt ribg d u h g c h h
AC
di qua diBm
M ( - 9 ; 2 ) . Tim tga dij cic Uinh ctia hinh chit nh4t
au 4 1
dikm). Mijt chikc hijp c 6 qui chu mPu
kg
q u i d u mPu d6 vP 2 qui chu miiu den. Chgn
ngiu nhien 6 qui ciu. Tinh xic s d t d&6
@
c$
duqc chgn c6
3
qua ciu d u
kg
qua c h miiu
66 vii
1
quh chu ,U u den.
lu 1 dikm). Trong kMng gian
v6i
he tryc tga dij
vu6ng g6c Oxyz cho cic d i h 1 ;; O),
B(3;
0; -3),
C(5;2; 6), D(0; -3; 1). Chimg minh&Ig c b di&mA ,
B, C,
D
1ii b6n dinh cha mijt
tir
di$n vii
tbh
th& ich
kh6i
tir d i b ABCD.
Cbu 6 1
diim).
ho
liing try dimg
ABC.A B C
c
ABCD.
lu 8 1
dikm). GiAi he phuong trin
Gu
9(1
dikm). Cho
a ,b,c
lii c i c s6 duung th6a m h
abc a c
=
b.
Tim gi i trj l h hht c6a bi&u hfic
NGUY NVANT H ~ N G
(GV
THPT
huyzn
L6
ujr D6n, Dd ~ 6 n ~
8/17/2019 Cac de Thi Tren Bao THTT
8/18
I I ~ ~ S ~ I C T R U ~ C K ~ T H I
ds68
77zGi gian I i rm bi i : 180phzit
3~
C u
(2 d i h ) .
Cho h h
=
+2
a)
Khlo sit
su
b i b thih v l v5 86
thj C )
cim h k
6
chq.
b) Vitt phuapg
trin ti& tuy&
cim d6 thj
C),
bikt
r5ng
ti@
tuyk 66 vu6ng g& viri dulmg thbg
dx+7v=O.
CAu 2
1 d i h ) . a
Gili p h k g
rinh
s i n x + ~ s i n ~ ~ - x ] = 2 .
b) Cho
s
phuc
z = 3
2i. inh m6 dun cim
s
phuc
z2
w=- -.
z+z
5
CClu 3
(0,5dih).
Gihi phucmg trin
2x4 21-x
= .
Ch 1 die ).
Giii bit phuang
trin
2 ( 1 - x ) J ~ 1 ~ 2 - 2 x - l .
dx
C u5 1
d i h ) .
Tinh tich p h b
I =
x ~ + x '
Gnu
6
1 d i h ) .
Cho
in
liing
tq
ABCA'B'C c
dhy 11
tam
giac diu cgnh
a, hinh
chih vubng g&
cim
A'
l h m t phhg t r k g v6i
t ia tam
gihc
ABC,
g& giiia m t
b6n (ABB'A')
v l mijt dhy
b h g
600.
Tinh th6 tich kh6i liing
t n~
BC.AIB'C
vA
khohg c h h giiia hai dulmg
h B
VA CC.
Ciiu 1 didm).
Trong mat phhg v6i he br(lc
Oxy,
cho hinh vubng
ABCD
c6
M -3; )
I trung diim
cim
AB,
dikm,
E,
thu& dow thhg
BC
sao cho
EC
=
5EB.
Bi6t r h g
DE
:
23x 9y
10=
0
vA dinh
D
c6 h o w dij dumg.
Tim
tga dij dinh
D.
CBu
8
1
d i h ) .
Trong kh6ng gian v6i he tga dij
Oxyz cho m t phhg (P): 2x
2y
z
5
=
0
v l
x-2 y-2 2-3
ulmg thhg
A :
1
1
2
.
Tim toa do
d i k A thu&,dubng
th hg
A
sao
cho khohg c h h tir
A
d h
m t
phbg
(P)
b a g
6.
C l u
9
(0,5 d i h ) .
Cho @phqp
E
=
{ l ,2 ,
3, 4,5 . .
coi ~ l l @ pgp,tht ccl chc
s
w n h i h
c
itnhht
3
chii
d,
hc chii si $6i mot k h h nhau th*
E.
Q g n
n&u nhih mi$
s
thu&
M
Tinh xac d t t
duqc chon c6 t6ng cac chii
s
biing
10.
CBu 10
1 d { h ) .X
cac
s
thvc khhg
iim a, b, c
t h b m h ditu ki
a+b+c=3. i
gil
trj
nh6
a b
c
nhht cim b i h thirc A =
- -
b4+16 c4+16
a4+16
P H A N v A N ~
GV
THPT c- Phan Bdi Chh , NghC An)
8/17/2019 Cac de Thi Tren Bao THTT
9/18
SUCTRU K~THI
obs 9
Thdi
gian
ldm bhi
18Ophtit)
Pu 1
(2 didm). Cho h h 6
y = m x 3 -3mx2 +3(m-1)
(m 11thamd 6 d8 thi (C,,,).
a) KhAo d t sv b i b t h i h va
ve
d8 thi ( C ) c h h h
s 8 t r h k h i m = l .
b) Chimg minh dmg d8
thi
(C,,,) lu6n c6 hai dihm
cvc
Ic
A va B v 6 i mgi m+O,khi d6 tim d c gid
Ic
c i a
m
d& 2ABZ OA2 OB2)
=
98.
Tinh gid tri c i a bibu thuc
b) Tim s6 ph3c z c6 mX un nhd nhit thda miin diiu
kien liz-31=lz-2-il.
C i a 3 (0,5 didm). GiAi phuung
trin
au 6
(1 didm). Cho hinh ch6p S.&C c6 day
ABC 18 tam gi6c vuting i
,
AB =3a, BC =5a; a t
p h h g (SAC) vu6ng g6c v6i mat i h h (ABC). ~ i k t
r b g SA = afi vb S= =30 .
Tinh
heo a th& ich
cua kh6i ch6p S.ABC vA k h o h g cach
tir
d i h
A
dkn
mat p h h g (SBC).
Citu 7
(1 didm). Trong mat p h h g vCri he tga do
Oxy, cho d u h g trbn ( C ) : ( X - ~ Y + ( ~ - ~ P = ~1
d u h g t h h g (A) : y 1=0 . T~ dihm t h u b
(A)
k
hai d u h g t h h g l
luqt ti
xcc v&i (C)
Qi
B vb C. im tga do dihm A bi&t ibg dien tich tam
g i b
ABC
b h g 8
C u8
(1 diim). Trong kh6ng gian v6i
he
tga d@
Oxyz cho mat c iu
(5 ):
2
3 z2 -a+@+ -22=0
vb mat p h h g
a)
2x
-
y 2 =
0
Chimg minh
r h g mat p h ~ ga ) cht mat chu 0he0 mot d u h g
~ s ~ ~ ~ X - J ~ C O S ~ X2 - s i n 2 ~ .
trbn.
X;Zlc
djnh t3111 v1 hn
kinh
c k u h g ron d6.
2
C i u 9
(0,5 dihm). Tim s& hpng khdng chua x trong
au
4 (1 didm). GiAi
he
phuung
trinh
khai tri6n
Newton c i a
*w
x9y
)
lu
I
(1
dihn). ~ h o lb c i c
sd
t h c d u m g thda
+ ' +3Xy(x-y)-12x2 +&=I
miXn x y . Tim gih Ic nh6 nhht c i a bi&u hirc
CPu 5 (1 diim). Tinh tich p h h
( x ~~-2)e~43+2x-*'+l&.
I j
,
(x+l)../=-
PHAM TRONG
THU'
8/17/2019 Cac de Thi Tren Bao THTT
10/18
ThWgian lbm bbi: 180phzit
Bhi
2 di&. Cho
h
JB
(SBC) v l (AX) l l 60°. Tinh the0
a
th&ich ~ 6 i
y=x3-3(m+l)xZ +9x-m C).
ch6p S.ABC v l khohg c k h tb t r~ ng
th G
clia
a ) ~ s h t v l v ~ d ~ t h j ( ~ ) ~ h h n s h k h i=0. tam g i b ABC d&nm@t hhg (SBC).
b) T m dk him 6 hai dikm CVc hi xl,x2 th6a
BU
7
(1 didm). Trong mat phhg v6i he t ~ ao
miin lx, -x,I=2.
Oxy cho in d n g ABCD co
t
l(1;4), dinh
-
3
BAi 2(1 dikm). An& tr6n d u h g thhg' c6 phuung trin
2x+y -1
=
0, dinh C nhm tr d u h g thhg c6
a ) ~ o< a < 1 , sina+fisin(;-a)=&.
phuung trinh x-y+2= 0. Tim tga
do
chc dinh
~ i n han
a
:
b) Cho phirc thda m h di&u ien:
2(z-1)=32+ (i- )(i+2).
Tim 12?--21.
BAi3
(0.5 d ih) . GiAi phuung
trinh:
loga (x
-2)
+
log, (x
-2)
=
log,
7
x1.
2
Bni
4 (1 didm). GiAibht phuung Mnh:
32x4-16x2 - 9 x - 9 J m + 2 > 0 .
u 5 (1 diim). Tinh tich p h h
A,B,C,D cfia hhh w&ngd l cho.
~ 8 ( l ' b ~ ~ w ~ i a n v 6 i ~ ~ t g a d 6xyz
cho&&n I(-1;2;3) vhm#tphhg (P): 4x+y-z-l=O.
~ i & tiuung Mnh m@t iu
th
I tikp xiic v6i
m t
phhg P) vh
tlm
tga do ti& dikm.
BAi 9 (0.5 didm). L p sd t,nh ih c6
4
chil s6 d&i
mot p h h biet
tb
chc chfi si5 0, 1,4,6. Tinh xhc s d t
d& 6 I duqc 18s6t, hi6n kh6ng chia h&t ho 4.
B i
10
(1 didm). Cho
3
s6 thvc ducmg x,y,z thda
1 1 1
miin
-+-+-
53. Tim gih trj,nhd nhht clia biku
y18 ~ 1 8
-
4 xU)lS
y2015 yu)15
zU)IS
z2015
x2015
I
=
I(3 cos3 i sinxbr.
thirc:
F =
ylw
y w
~ w zlm
.
0
U 6 ( 1 &).chohlnhch6p S.ABC
c
AB=AC=a
BC
=
30°, SA ABC), g6c gifla hai mat phhg
NGUYEN V N X 3
(GV
WPT Ygn Phong
S
2,
B ~ C
inh)
8/17/2019 Cac de Thi Tren Bao THTT
11/18
ol s6 2
-
,
ntdi
ian
lcim Wi: 1610phlit)
*
2 x
-
Clu
2
dihm). Cho h h 6
y
=
C A U ~1 46111).
~ h o m
@Bc ABC vu6ng $i
x
A I I ~ ~ ,
pnh
BC
a5 phu-
a
+ 3
=
o
V
a ) ~ h a o d t w b i h l h i h v l v e d h t h i ( c ) & b d .
& ~ ( ~ ~ ) .
b) ~ i & thuung trinh ti& tuy& cha dh
t
(C) tqi
dikm
~ 0 ; - 1 ) .
ciic d o ~
D,CD Tim
tog dij clia
B,C
bi&tdubng
trbn ngogi tibp
tam
g i k
DEF
di qua dikm
CAut(ldi&).a)C oalig&cmAmta=2.Tinh
M Z:-l+a).
msa
P =
Clu 8 1
dihm). Cho m t phhg
P):x-2y+2z-1=0
sin3a+3ms3
a
b) Tmng m t phhg
Oxy
tim
t
hqp c8c d i h biku
CC
duhg
X - 1 y -3
z
X - 5 y z+5
di8ndphhc z thoAmh lz- 3-4i)1=1. dl
:-=--
-3
-2
d 2 : - = - = - - .4 -5
Caw
3
0,5
dihm). GiAi phuung trlnh
Tim dikm
M
thuijc
dl, N
thuoc
dz
sao cho
M N
song
1 8 song v6i
(P)
P
~ u b n g hhg M N c b h (P)m ~ t
-loga
X 3 )
+-log, X -
) = og24 ~ .
2
k h o h bibg
2.
clu 1
didm).tho
0 <
.
c h h g
hg
~ U 0.5
didm). Gihi b6ng dB nil v6 djch D b g
v6i mQis6 thvc
a , c6
Nam
A
n h 015 duqct6 chhc tqi thhh ph6
H
Chi
Minh
c6
dai Mng tham
dy.,
trong 66 c6 hai dai
x2-a2
a 2 - y 2
2 ~ - ~ )
->-.
VietNam vP Thgi Lan. CBc dQi b6ng duqc chia n g h
x 2 + a 2 a 2 + y 2 - x + y
n h i h
thBnh 2 b h g c6 s6 dQi b6ng bibg nhau. Tinh
2 xiic suht sao cho hai dai Vie Nam vl Thiii Lan nhn
cru
1 d ih ) . ~ i n hich phh:
= ~ x - 2 ) E k
i b h g ~c
u.
Cbu
1
1
di&). Cho
a
18
sh
thvc duung. Gir3i v l
Clu
6 1
dihm). Cho hinh ch6p
S.ABCD
c6 diiy
18
x
-2 x
hlnh vu6ng cpnh
a,
m t
b h SA B 18
t m gihc
b i b U@ PhumgWsau
= a
x2 -1- J i
dhu,
sc
=
SD
=
a a . Tinh
thk tich kh6i ch6p
S.ABCD
K&u D ~ HINH
vticosin
6c
gimhai m t
p l h g
(SAD)vti SBC).
GV THPT chuygnHting V m g ,
Phli
Thp
8/17/2019 Cac de Thi Tren Bao THTT
12/18
~ ~ C T R U fmr
s6
3
(The gian lam bhi:
180
phlit)
CBu (1 didm). Khko sit sv bi&nhien v i vE db
t h i h h s 8 y = ~ 3 - 6 ~ 2 + 9 ~ - 2 .
Ciu 2 (1 A).
Tim
gih trj hnhh, gih trj nh6
d t a lun
s
y = 3 ( G i + f i ) - Jm
Cau 3 (1 diim). 1) Tim @p hqp chc di&mbi&u
dign bong mat phhg phtrc ciia s8 phtrc z sao
cho lzl=I;-3+4il.
x 1
2) Gihi PT 82) -'
=
8 ) (x W)
CBu 4 (1
d i i ) .
~ i n hichphin: I = j E d x
x2
.
2
CAu 1 di&). Trong khang gian
v6i
he tga do
Oxyz chom cb 9:
x-J)2
+0+2)2 +(z-v =lo0
v i
mat phhg (P):2x-2y-z-9=0. C h h g minh
CPu
7
1 dibm). Cho hinh 1hg tn
tam
giac
ABC.A'BtC' bi&
AB=q
AC=2a
v i
=C= .
Hinh chiku vuhg g6c c6a A' lbn mat
p h h g ( ~ ~ ~ )riing v i trgng ttim G c6a
tam
giac ABC, g6c giitaAA1 va A'G bkng 30'.
Tinh th& ich kh6i lilng v i khohg cich tir
C' dhn mat phhg (A'BC) theo a.
Ciu 1 d i h ) . Trong
m t
phhg v6i he tga do
Oxy, cho
tam
giic ABC c h gi A. Cqnh BC
c6 pbucmg
trinh
2x y
+
1= 0, d u h g cao ha
tir
dinh B c6 phuung trinh x+3y-4=0 v i di&m
H(1;4) nhn trbn d u h g cao ha
tir
dinh C. Tim
tga do chc dinh ciia
tam
giic ABC.
Cia 9(1 d i h ) . Gihi bht phuung trinh
x(x+~)(x-3)2&+=-3, (XEW).
r b g (PI ckt
0.
hi 66, tim
,
V h
cii~ 0(1di4m). ~ O , y, z 18 ciic s8 thvc
clia d u h g tr6n
this
dien c6a(P) cat 9.
Cnu
6
(1 didm). 1)Cho
tan
a = 3. HZLy tinh
duung thba man i
= 1. Tim gia
8cos3a+4sin3 +3cosa
tr
nh6 nhht ciia bi&uhthirc
B =
P=
x3
+ y3 +
z3
2cosa-5sin3 a
2)
Tim
sd
hqng
%tmg chinh
g i b
bong
khai
trign
2(2+x)y 2(x+y)z 2(y+2)x '
N G ~ N
UANG
THI
(GV
T PT
Brjo U c ,Ldm
ng)
8/17/2019 Cac de Thi Tren Bao THTT
13/18
Tm]rsirc~~u c~m
L
sd
(ZWigian
lhm bhi:
180phut)
Ciu
1 (1 dikm . K h b sht
SIJ
bikn t i v1 vP, dB thi
2x+1
c l h ~ s 6=
x-2 '
CAu
2
(1 diim . L p phuung trinh d u b g thhg A
cht h h b y=x3-2x2-x+2 tqi hai di6m
A,B p h b bikigt sao cho AB = rong d6 Ui6m A
thuoc @c tyg.
Cda 3
1
dfinr . a)Tim hkg
d
hdngc h h x trong
khai
IS
t r i & & t h i ~ ~ ~ e w t o n c l ak - n 6 ) ,x>O bikt
dng
t t
he sd tnmg khai
tri
nAy k
( ~ E W ) .
b) Gihi phuung
trinh
3l0s2 32-10g2 = 10.
C ~ U1 ).inh gibi hqn
I
= lim
ex- cosx
x o
x
C h 1 dl ,). a) Cho s6 thvc
a
thba miin
cosa-sina=-. Tinh A=tana+cot2a.
J3
b) Mot hohg tir di slln
th6
vbi xhc suit bib tding
thb @
lkg nghjch v6i khohg chch tir hohg
tir
dkn
thb. d khohg chch 20m khA n h g bib tding c6a
hohg tir 11 50%. N U
3
khohg chch 20m hohg tir
bib khdng tnhg thi hohg tir sP,bib ti@ l h hu 2 (6
khohg chch 30m) vA nku l h 2 hohng tir
b h
kh6ng
tmhg thi hohng
tir
s&bib tikp l h hir 3 (b khohg
c k h 50m). Tinh xhc s d t d6 hohg tir bib tding
th6
sau n h i h nhh 3 l h bin.
CAu
6 (1
).
Tmg khbg
gi n
v6i M W t9a d6
oxyz,
w ch (S1 ,(S2) a p h m g trinh
l n
Chimg
minh
ribg (S,) v1 (S2) cht nhau the0 giao
tuy€n 11mot d u h g trbn (C). Tim b h
kinh
v1 tga
dot c k (C).
Cia 7
1 4th .Cho hinh ch6p S.ABCD c6 dhy 11
hinh thang vudng tqiA,D; AB=AD = a vA
CD=2a. ~ i & t&ng hai @t phhg (SAC) v1 (SBD)
c b g vudng g6c v6i @t phhg (ABCD) v1g6c giiia
hai mat phhg (SBC) v1 (ABCD) b h g 45'. Tinh
the0 a th6 tich kh6i ch6p S.ABCD v1 khohg chch
giea hai d u h g thhg SD,BC.
C h 1 A).
Trong mat phhg vbi hkg trpc tga do
Oxy, cho tam gihc ABC wdng
tqi
A ngo@i tikp
d u h g trbn (C) t n K c6 D A ti& digm c6a (C)
tren qmh AC. D u b g trbn ngogi tikp tam gihc
BCD cit c& AB
i
i6m E khhc B. Chc d u h g
thhg
qua
A,D v1 vudng g6c vbi CE c6t c d C
Qi F v1
G.
Tim tga do chc dinh c6a
tam
giQ ABC
bi&tF(-3;-4);G(1;-1) v1 K(-2;3).
Cia 9 (1 I). GiAi bht phuung trinh
~ ( x - ~ ) J ~ + ( x + I ) J ~ < ~ x - ~xEW).
C u 1 1 ddm .
Cho chc s6 thvc duung x,y,z thba
m b di€ukikign
- - -=-
I I lo . T h g i h h i l h
x .
y
z x+ y+ z
nhht cua bi6u thirc
GV
HPT huygn
Hri
Rnh)
8/17/2019 Cac de Thi Tren Bao THTT
14/18
8/17/2019 Cac de Thi Tren Bao THTT
15/18
. L
U ~W ~W K ~T I
DL s6 s
(TA
gian lirm
Mi
80 phdt)
1 ) Khtio sht sg bikn th ih vh vt 86
thi
(C) d a
hhm
s
dti cho.
2 )
Tim tham s
m
d&,duhg hhg
d:
y =
x
n
&it db thi
(C)
Mi
2
di&mp h h bigt
A, B
sao cho
tam
gihc
OAB
Mlang tqi
0
0
8gbc tQa do).
2 )
Tim gih
tri
16nnhk v l nh6 nhht c6a h h
y = J F i Z + J 4 - x 2 .
1)
Cho
s
phirc
z
hda
msln (I-2i)(z-1)
= 2+3i)z
Hiiy tinh
z.;
2 )
Tdn gib shch c6
3
ngh, n g b
thix
nhht c6 5
cuhn shch g i h khoa To&
vA
5 cubn shch bai
@p Toh; n g b thu hai c6
6
cubn stich
giho
khoa
Ly
va 6 cubn sbch Mi @p
Ly
v l ngiln thu
ba co cubn shch &o khoa H6a va c u h
dch b&
@p
Hba.
L
luqt lhy nggu nhien m6i
n iln mot cuhn, tim
xiic
suit d6
3
cubn 1hyduqc
i
deu 11shch bai @p.
C k 5
(1 km)
Trong khdng gian
Oxyz,
cho
x = l + t
d u h g thing
A : y = 1+
va
it
phhg
z = l + t
( P ) : x - 2 y + 2 z - 3 = 0 . Tm @ @
dikm
t n
m c tung
s o
cho kh&g &h tir
ih
d6 d&
duh g hhg
A
b h g k h h g chch
tir
d i h 86 d b
a t hhg
P).
Ch 1 d m )
Cho 15ng
ABC.A 'B'C'
cC
A 'ABC
18
t r
di dku c&
a.
Tinh theo
a
th&
tich khbi t r dien A
'B'BC
v l khohng chch giila
hai d u h g hhg
AB
va
C C ' .
lu
7.
1
d i im)
Trong m#t phibg
Oxy,
cho
d u h g trbn
( C ) x2 y2
-
x -
y =0 Tim tga
dij chc di6m
A,B,
nhm tren d u h g trbn
C)
sao
cho
OABC
lah i hiinh@t 6
AB
=
2BC.
Can
8. (1 dihm)
GiAi he phuang trinh
* +,
(1 dibm)
Cho s thvc x, y, z th6a miin:
x 2 + y 2 + z 2 = 8
Tim gih
tr
16n nhht va nhb
xy yz w=-4
-
3 x3
+y3
+ z 3
in
2x
nhht cGa bi&uhirc
P
=
u
4 5 i im)
Tinh tich p h k
=r d ~ ~ + ~ ~ + z ~
ax3
8/17/2019 Cac de Thi Tren Bao THTT
16/18
, ,
~'f.' P
,: ?
PI
s6
7
(Zk ian l im
bii
18Ophdt)
'+
g
1
didm).
KhAo sit v i vv~.B
thi
h m
sb:
phhg qua
M
v i vuBng goc vol
d
Tim
rpa
ap
. -- y = x3 + ( 4 m + 2 ) x 2+ (m 2+ 8 ) x + 2
AABC v i ASAB i c6c tam gihc dbu, hlnh chiku
r ;
% 3
vu6ng g k h
S
Sen ABC) 18 trung dibm clia
A B , s c = ~ , pi M l i tnmg di&n ciia SC
a
'sg
ph
z
ikt
z
c6 p h h ~huc v i p h h
Ao
~ i n h tich ch6p
SMC
v
w~~~o l
d6i nhau,z 6 m o c k b h g modun c b
c6ch gim AMv6i
SB.
I
w =(1+2i)(l-i) .
CPu 8
1
didm).
Trong mat phhg v6i he
Qa
do
b) GiHi phucmg
Mnh: 32.4x
=
1.p 27
Oxy cho W C h qiA H lk trung d i h
I
3-la 4 1 didm).
I&&
BC,
D
l i
h nh
chiku vu6ng g6c c bH t n AC
1
a) Cho
sinx+cosx=-. Tinh
l i trung b i b c hED hamg trkh
2
A
=
2sia3x.cosx-sin4x.
dow thhg nhi h i dinh bit ki ah
a
giQ
(H).
TiF S
chpn
2 dom
thing
bht
ki tinh
xhc s d t 86
trong hai
dam
thbg duuc chgn c6 it nhit mot
dozp thhg l i cqnh c b a gihc H).
CBu 1 didm). Tinh tich p h h
I - 1
X
-
t
dtc
sin2x
.
-
r -T
eael6
1 d i k ) . Trong kh6ng
@an
viri he tpa 8
nyz
cho d i b M(3;-1;-2) v i d u h g thhg
d
x-1 y + l
z
--
--=-
. vi phucmg m t
d u h g
thhg
BD:
x +
y 4
=
0:
p~ungtdnh
d u h g thhg
AB:
3x + y
-
10 = 0. Tim Qa dij
di6m
C .
ChQ (1 dikm).
Gili phumg trlnh:
;i
xS -7x2 +9x+12 r l I
= ( ~ - 3 ) ( x - 2 + 5 ~ x - 3 ) ( - - 1 ) '
Q 0
1
dGm).
Cho
ba s6
thvc duodg
a
b,
c
I
th6amh a + b + c l 2 .
Tim gih
tfi
nh6 nhht cda:
dc
P=
lAb +
c 1
+-
a+ +b+b+ +c ~ + + a 2 J a . J
~ r
8/17/2019 Cac de Thi Tren Bao THTT
17/18
ot
s6 a
(77113gian lbm bbi: 180
phtit)
.
(1 dikm). K h b sht sv bihn thien va vC
b) Mat b@d&
thi ~ Q C
inh gi6i vbng truhg ma
~dtlqhhms6 = ~ 4 - 2 ~ 2 + 4 .
m6i d&gbm 5 c hduqc c h ~ nir 15ciu &, 10ciu
trung
binh
v i 5 c&u kho.Mat dk thi duqc g ~ i8
f l u
2
1 didm). im gi6 tr 1 n nhbt v i gi6 tr
, , ,
4
'tot nhu trong d&hi c6d a ciiu d8, trung binh
mo nhht cca him s6
f
(x)= x- rsn
va kh6, dbng thiri s6 ciu
&
khdng it hm 2. ~b~
x-1
neb n h i h rnijt & thi trong bijd . Tm xAc
foqn 12; 51.
silht d dk thi lby ra lamat d8 thi'dt .
Clu
3 (1 bdm). a) Cho
s
phuc zth6a miin
-
7 (1 didm). Cho t x dien OABC
v i
OA=a,
(2
-
3i)z
+
(4
+
T
=
-(I
+
4i)2.
im
phhn
thvc
ph&I o cGa
z
, , OB=~ , =C, ATB=U, Z?=45~ ,
C~A=W.
.--
3)
Giki phuung trinh
+F
Gqi
CH
va CK I& luqt la dwhg cao aia t m
6
gik OACvhOBC.Tinh th&ich tir dienO m .
10g~~(x+ l )2=10g 2 4 4 - ~ + l ~ g s ( ~ + 4 ) 3 .
4 ~8 f1 didm). Trong m t phhng v6i he tga dij
(1 dikm). Tinh tich p h h
p.
.I
Oxy, cho h i udng ABCD c6 M lh
0
cosx- inx s - 2
4
diim cqnh BC, phuung trlnh duhg thhg
I =
J
sin&
+
2(sinx cos
*&kdr.
0114:~-y-2=O,dinh C(3;-3)va dinh
A
nim
--
4
I = ~ tren d u hg thing d:3x+y-2=0.
X b
d m
CHU5 (1 dikm). Trong khbnpgian v&i he @adij
t ~ ao di&mB.
O ~ Z
ba
d i h
A(1; 1;
11, B(-1; 0 21,
9
(1 bibm). ~ i h i
phwng
C @
-1; 0). im
tga
do
~ t i h
tr tia Ox
iao cho th& ich kh6i tix
dipl
ABCD
b b g
2 Mi
+1+1 (X2-Y3+3y-2)
(x,yEN.
16h5y vibt phuung trinh mat gogi adp~ +y2)2 U)15y2+U)16=2$4032~
.
iienABCD.
6
:='
16(1 didm). Cho a,b,c [Q 11.Tim gih
tr
3hr 6 (1 dm). ' 16n&bt
c h
bi&uhirc
1 Tinh gi6 trj cua bi&uhirc
P
=abc+(1-a)(l-b)(2-c)
2015cos2a+2016cos2 - - - - 1 1 a b
- c
2
P =
l + d l + d b+c+l c+a+l a+b+l
1 sin2a.sina
3 PH f TIRQNG
rn
8/17/2019 Cac de Thi Tren Bao THTT
18/18
&: : --
**Y+?I
~e s6
Thdi
gian Icim
brii:
I8Ophlit)
(1 ~di m).Khao t 8 y bikn thien va
v t
d8 thi
b) C6 40
dm
he &rqc d h l h uqt tl d& 40.
3 2x+3
Chon ngh nhi6n ra 10dm the.
Tihh
xirc s& d&rong
@ h b s $ y = -
x + l
'
10dmth6duqc chQnrac65&th6mang
s
16,5dm
a ~ ( l & ) . v i & p h u m g ~ t i ~ w & d d b ~ t h ~ m ~ d c h b t r d n ~ d 6 c b i c 6 d h ~ m @ t b t h e
h h d y=x3-3x2 +x-l ,bikt t ib tuyhd6c6hed
mang s8chia
hkt
ch 10.
C @ (1 d&). Cho
hinh
ch6p
~
gih S.ABCD c6
k h & g d h t i r ~ ~ & ~ d & @ M & g a . ~ i n h
(1dikm).
@eo
a t44 icix,khi i
h6p
S.ABCD,
va
khohng
d c h
gjih
a)
T r h
a t , hhg phlrc,
tim
t$p hhp cc8c diih b i b
h
D,
SC, bi&
hj ia -g
e o
dihdphirc'
r m6arntin
Iz-l+il=2.
hinh
ch6p S.ABCD vA m#t hg 30 .
b) ~ o ti$i
ihy n h h
m g
i@ cijt
d i b bh dhu
15
m6c
~ g lb
1 A .n g wt
v6i
%t~i
O ~
gi i
cays63d & m & g i 6 i c ~ y d 6 6 m ~ c o n d u ~
ho
o hg
BCD
vuemg
i
A va
D
tinh 4. cir
300 met dwg mot a t dib. H6i c6
tit CB
AB =2AD,CD =3AD.
Mg
BD 6 pbuong
bao
n h ib
@t @I&rqcdm
A
trinh x-2y+l=0 , du6ng &kg AC di qua Uh
,,
'
d 4;2)
Tim
tp
'84-
dinh
A
bi&
dng
&@Itich
hinh
~ &m) . ~ i n h d c h ~ h i l u= Isin&&.
o
th ng BCD
b j 10vadi&
A c
h&
hun
2.
djl
1 &&).
, ~ f ;ng fiag g i m v
he
6
o EY.9 (I di&m)). m
dt
c i c k
gis
t i cha tham s8
- ~ ) ~ ~ ~ . c h o , ~ t ~ h h ~i P ) : x + y - 2 ~ + 1 ~ 0 ~ ~ m * c h o h e s a u * n g h i b.
v&0 < a e E . ~ i n hihtri c h i b thuc
2
~ & x u h D q I
(CV
THPT hy2n Vinh PMc)