Cac de Thi Tren Bao THTT

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


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 .


3/18


4/18


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)


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 .


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 ~


8/18

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)


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'


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)


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


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)


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)


14/18


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


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


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


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)

Cac de Thi Tren Bao THTT

Documents