tuyen chon 500 bai tap toan 10 nxb ha noi 2005 2 885

B a i 287. (Cfiu hoi trftc nghi^m)

b)

,

FhiTdng trinh (1) »

Phi^cfng trinh nao sau day c6 hai nghi$m X i , X2 thoa m a n bit't d^n
thii-c xi < 0 < X2 < 2.
a)

3x^ - 5x + 1 = 0

b)

3x -

.2

X

t = x^ - 2x + 2

(x' - 2x + 2)^ - 4{x'' - 2x + 2) + (2m + 4) = 0

(dieu k i ^ n : t >. 1)

- 4 t + 2 ( m + 2) = 0

c) Sx"* - 2x - 4 = 0

(*)

D a t fit) = t^ - 4t + 2(m + 2)

d) 3x - 5x - 4 = 0

+4 =0

PhUOng t r i n h da cho c6 n g h i e m x c=> Phifcfng t r i n h (*) c6 n g h i e m t > 1

• HUdng dan
D a t f(x) = ve t r a i phuong t r i n h bac h a i
Ta p h a i c6 :

B i e n d6'i phuong trinh (1).

f3.f(0) < 0

ff(0) < 0

3.f(2) > 0

f(2) > 2

Trifdng h<fp 1 : PhUOng t r i n h (*) c6 h a i nghiem t i , t2 > 1
S •
-

^

V, ,

-2

.
t2

Dap so : 3x^ - 2x - 4 = 0 (Cau c)

B a i 2 8 8 . (Cau hoi trie

nghi^m)

Vdi nhffng gia tri nao ciia a thi phi^c^ng trinh 3x^ - ax + a = 0 c6 hai
nghi^m X i , X2 thoa man - 2 < X i < 2 < X 2 hoSc
a ) a < - 1 2 V a > 4
c)

d)

A' = 4 - 2 ( m + 2) = - 2 m S 0

l.f(l) > 0

l.f(1) = 2m + 1 > 0
I

Xi<-2

b ) a < - 4 V a > 1 2

-4
A' > 0

•

-12
= 2 > 1

t i < 1 < t2 o

(hien nhien)

l . f d ) < 0 CO m < -

Hofp h a i k§'t quA -

T a p h a i co f ( - 2).f(2) < 0

2

Trufofng hcjTp 2 : Phuong t r i n h (*) c6 h a i n g h i e m t i , t2 sao cho f

* Hiidng dan
D a t f(x) = 3x^ - ax + a

<=>

2

- < m < 0 V m < - - t a c 6 m < 0
2
2

Dap so : m < 0

Dap so : a < - 4 V a > 12 (cau b)
B a i 290. (Toan ttf lu^in)
B a i 289. (Toan ti^ l u $ n )
a)

Tim g i a t r i n h o n h a t ciia h a m so' y = x^ - 2x + 2

b)

Cho phtfoTng t r i n h (x^ - 2x + 2f - 4(x - l)'^ + 2m = 0

* Hiidng ddn
Bien ddi y (

b)

Dat t =

Tim gia tri Idrn nhat ciia ham so y = - x"* - 4x'* + 5

b)

X a c dinh a de phtftfng trinh (- x'* - 4x'' + 5)^ - 2(x'' + 4x*) + m = 0 c6
nghiem.

(*). X a c d i n h m

de phifoTng t r i n h (*) c6 n g h i $ m .

a)

a)

f + hkng so' A > A

*Hudngd&n
a)

B i e n d o i y = h ^ n g so' (

b)

D a t t = - x"* - 4x^ + 5 r o i b i e n d o i phiTcfng t r i n h da cho t h ^ n h
phirong t r i n h bac h a i a \o t (dieu k i ? n t <

r o i b i e n doi phiTcrng t r i n h (*) t h a n h phifcrng t r i n h baC

hai in so t (dieu k i e n ciia t : cau a)
Phuong t r i n h da cho c6 n g h i e m

<=> Phuomg t r i n h (theo t ) c6 nghiein

t >

...) va liAi y x**, x^ > 0

Sau d6 g i a i gio'ng b a i 289.
Dap so :
b)

max(y) = 5 v i y = 5 - (x" + 4x^) < 5 (Dfi'u = xdy r a o
R

,

GlAl
a)

T a c6 y = x^ - 2x + 2 = (x - 1)^ + 1 > 1, Vx £ R
Dau = xay r a o

x = 1

Dap so : m i n ( y ) = 1 k h i x = 1
R

166

).

c)

Phucfng t r i n h da cho o>

x = 0)
•

t = - x"* - 4 x ^ + 5

(t < 5)

t^ + 2t + m - 10 - 0

(*)

*

T i e p tuc g i o n g b ^ i 289, t a c6 : - 25 < m < 11 V m < - 25 <=> m < 11
167


E.

KI^M TRA CAC KIEN THLfC VE TAM THUTC BAC

HAI

a i 294. (Cau hoi trac nghiem)
Gia silf tam thtfc bgc hai f(x) = (1 - m)x^ + 2mx + 4 c6 bang xet dau :

B a i 291. (Cau h6i trSc nghiem)

(xi, X2 la hai nghiem ciia f(x))

Tam thtfc bglc hai f(x) = (m^ - 3)x^ + 2mx c6 bang xet dau

+

+00

1

0

f(x)

0

-

0

+

c) m = 3

d) m = - 3

b)m

4[ HUdng ddn

• HUdng ddn

+

0

-

c)lml>l

d)|m|
f(x) > 0 k h i x e ( x i ; X2) n e n a = l - m < 0

c:>m>l

Dap so': m > 1 (cau a)

TU bang xet dau t a c6

f(l) = 0
- 3 > 0

B a i 295. (Cau hoi t r i e nghiem)
Tam thiJc f(x) = 2x^ - ax - 3 c6 hai nghiem X i , Xg thoa man dieu

Dap so : m = - 3 (cau d)

ki^n —

+ —

B a i 292. (Cau hoi trfic nghi^m)
Xac dinh cac gia tri cua m de bat phifoTng trinh x^ - 4x + 2m - 1 < (I
CO t^p nghi^m S = 0 ?
5
a)m<—
2

0

C a u nao sau day dung ?
a)m>l

b) m = - 1

+ 00

X2

-

f(x)

Hay tinh m ?
a) m = 1

— 00

X

- 00

X

5
b)m>—
2

5
c)m<—
2

d)m>

5
—
2

<=>

a) a = 15

= 5. Tinh a ?
b) l a I = 15

c) a = - 15

d) MOt gia tri khac.

* HUdng ddn
1

1

X-^

X2

1

2

S

P

Xj •X2

Dap so : a = - 15 (cau c)

* HUdng dan
Bat phifcfng t r i n h da cho v6 n g h i e m

X.,

- 4x + 2 m -

1 > d

n g h i e m diing vdi m o i x E R <=> A' < 0

B a i 296. (Cau hoi trac nghiem)
Tam thii-c f(x) = x^ - 2mx + 4 c6 gia tri nho nhat bSng 3. T i n h m ?

Dap so': m > — (cau b)
2

a) m = 1

c) I m

b)m = - l

I = 1

d) I m

I >1

* Hudng ddn
B a i 293. (Cau hoi tr&c nghiem)

•

T|lp nghiem cua b a t phufoTng t r i n h

_

2x - 1

>0 1a :

(x + l)(-x^ + 4x - 6)
a)

S = (- 1 ; ^ )

b)

S = [- 1 ;

1]

c) [- 1 ; ^ )

D a p s o : S = (- 1[ ^]

(cau d)

> 4 -

Vay minff(x)] = 4 -

= 3

Dap so : I m
d) (- 1 ; ^ ]

• HUdng ddn
Vi {- x^ + 4x - 6) < 0, Vx e R n e n bat phiTcfng t r i n h da cho tiTOng
2x — 1
1
duong vdri
<0
<=>-l2
*

f(x) = (x - m)^ + 4 -

B a i 297. (Cau hoi trSc nghiem)
Ham so y = ^2x^ - m x + 2 c6 tgp xac dinh S = R k h i :
a)

I m l <4

b)

I m l <4

c)

I m I < 16

d)

I m I < 16

* Hudng ddn
H a m so da cho xac d i n h t r e n R o
^ a p so : I m

168

I = 1 (cau c)

2x^

>o,

VxeR'<=>A<0

I < 4 (Cau a)
169


B a i 298. (C&u hoi trfic nghiem)
Cho tarn thtfc f(x) =
f(x + 1) = 0 la :
a)S=|l;3)

- 4x + 3. T|lp nghiem ciia phi^Ong trinh

b)S=|-l;-3|

c) S = |0; - 2|

d) S = {2; 0)
'

- 2

f(x) > 0, Vx

0

(f(x) > 0, Vx
X

+00

B a i 299. Cho a, b, c la dp dai ba canh cua mpt tarn giac.

D h a m so

+

- 00

E

R

D = R

e

R

D = R

Xi

+00

X2

(-00 ; x i j u fx2 ; +oo)

f(x)

x i , X 2 l a hai n g h i e m ciia f(x)

ChiiTng minh rSng tarn thrfc f{x) = (b + c)x^ - 2(a + b + c)x + 2(a + b + c )
ludn C O gia tri dUOng.
• HUdng

Xet dau fix)

-

g(x) = f(x + 1) = (x + 1)^ - 4(x + 1) + 3

Dap so': S = (2, 01 (cau d)

•

A'

- cx;

% HUdng d&n
•

m

Xi

=

-

2 - 7m + 2

Xg

=

-

2 + ^m + 2

(xi <

X2)

dan

Ti'nh A' cua f(x) va chii y r ^ n g a, b, c > O v a a + b > c , b + c > a ,

GlAl
•

f(x) = (b + c)x^ - 2(a + b + c)x + 2(a + b + c)
A' = (a + b + c)^ - 2(b + c)(a + b + c)
= (a + b + c)[(a + b + c) - 2(b + c)]
= (a + b + c)(a - b - c)

,

V i a, b, c la do dai ba canh ciia m o t t a m gidc nen a, b, c > 0
A' < 0

a < b + (

f(x) > 0, Vx € R

B a i 300. T u y theo m.'tim t§p xac dinh cua ham s6' y = ^x'^ + 4x - m + 2 (1)
• HUdng d&n
•

H a m so (1) xac d i n h o

x^ +

> 0

•

D a t f(x) = x^ + 4x - m + 2, t i n h

xet dau A'.

GlAl
•

H a m so (1) xac d i n h o

x^ + 4x - m + 2 > 0 (G9i D la t a p xac dinh

cua ham so)
•

D a t f(x) = x^ + 4x - m + 2, t a

"5,

CO

m + 2

-2

- 00

170

: A' = 4 - (- m + 2)

0

+

171


VECTO

Chu^'oii d o I
Kicn
I)

t h i f c coT

c)

*
->
Do dai cua vecta AB la dp dai ciia doan thang AB, ki hieu |AB|
0 CO do dai bang 0 ( 0 = 0 )
3) Hai vector bang nhau
Dinh nghla
Hai vecta bang nnau khi chiing ciing hudng va c6 dp dai b&ng nhau.
Chii y :
• a va b bang nhau, ki hieu a = b

ban

- >

C a c djnh nghTa

1) Vector
Dinh nghla
Vecta la doan thSng da dinh hirdng, nghla la da chon mot diem mut
lam diem dau, diem mut con lai la diem cuoi.

•

Vdi hai diem phan biet A va B, ta c6 hai vecta khac nhau :
AB va BA
• Vecta CO diem dau va di§m cuoi trung nhau, chang han: AA , MM ,
goi la "vector - khong", ki hieu 0
.
,
'
2) Phifofng, hi^oTng, dp dai ciia vectof
a) Dinh nghla
Hai vecta goi la ciing phuang khi hai vecta nay Ian liicrt nhm tren
hai dUcJng thang song song nhau hoac trung nhau.

H$ qua : Hai vecta cung phuang vdi mot vecta thuT ba thi hai vecta
do cung phi/ang.
b) Hai vectcf a va b cung phifcfng thi a va b c6 the cung hxidng
hoac ngUdc hi^drng
Chii y :
• 0 cung hudrng vdi moi vecta.

• hirdrng.
Hai vecta cung hudng vdri mot vecta thiJ ba thi hai vecta do cung
—>

172

a = b va b = c

doan th^ng
vecta
Vecta CO diem dau la A, diem cuoi la B, ki hieu : AB

•

Dinh nghla

>

a = c

• Cho san a va mot diem O, ta c6 duy nhat mot di§m A : OA = a
• Mpi vecta 0 deu bang nhau
II)

Phep cpng c a c vectd

1) Dinh nghla tong cac vectof
Cho hai vecta a va b .
Tii diem A tuy y, ve AB = a va BC = b .
Vector AC dUdc goi la tong hai vector a va b , ki hi^u : AC = a + b

Chii y :
• Tong a + b khong phu thuoc vao vi tri diem A.
•

Quy tac ba diem : AC = AB + BC (ba diem A, B, C tuy y) (hinh 1)

•

Quy tac dudng cheo hinh binh hanh.
ABCD la hinh binh hanh =5 AB + AD = AC (hinh 2)
A
D
B^
(Hinh 1)


2)

Tinh chat
->

^

a)

V6i moi a t a c d :

a + 0 = 0 + a=

b)

V d i m o i a , b , ta c6: a + b = b + a

c)

Vdi moi a ,

b, c,tac6:(a

(giao hoan)

V e c t d d o i c i i a mpt vectof

a)

Dinhli

a +
b)

X

=

b)

k. a = 0 neu k = 0 hoac a = 0

+ b) + c = a + (b + c)

(ket hcfp)

c)

Do dai ciia k. a la |k. a | = i k | . | a |

2)

Tinh chat

cho trUdc, lu6n c6 m^ot vecto

x

duy n h a t sao

cho

0

Dinh nghia

a)

k ( l . 2 ) = (k.l). a

b)

(k + 1). a = k . a + 1. a

c)

k( a + b ) = k. a + k. b

d)

1. a = a ; 0. a = 0 ; k. 0 = 0

3)

Dinhli
->

Neu vecto

a

+ b

= 0

—»

t h i vectof

b

dirge goi la vector doi cua

so' thiTc k sao cho b = k. a

d

a + (- a ) =
M 6 i vector c6 m p t vector doi duy n h a t .
•

-»

->

k > 0

•

k < 0

Dinh nghia
Cho h a i d i e m j h a n biet A va B

b)

,

M chia doan t h ^ n g A B theo t i so' o
c)

= a+(-b)

1-k

OA -

^

k.OB

K h i k = - 1 t h i M la t r u n g d i e m ciia doan t h i n g A B .
Vay :

-

OM =

(k / 1)

H§ qua

Phep t i m hieu a - b goi l a phep trir hai vectcr.
>

cx> M A = k. M B

Dinh li

Dinh nghia

-

'

D i e m M chia doan t h ^ n g A B theo t i so k
Cho d i e m O tCiy y va k ?i 1 , ta c6 :

>

> ,.
'

H i ? u c i i a h a i vectof

-

o a v a b cung hddng
-*
—>
o a va b ngupc h d d n g

a)

Chiiy:

1)

•

D i e m c h i a d o a n t h a n g theo ti so' cho trifdfe

N e u b l a vector doi cua a t h i a la vector d o i ciia b n e n :

va vectd doi cua vector b nghia l a : a - b

IV)

->

0 t h i t o n t a i duy n h a t

4)
-»

Hi#u cua vector a va vectcf b , k i hi^u a - b , la tong cua vectcf a

•

M la t r u n g d i e m ciia doan thSng A B
o

>

OM =

If

OA + OB

( 0 la d i e m tuy y )

Cho ba d i e m bat k i : A, B, C, ta c6 : AC = BC - B A

Phep nhan vectd v6l mpt so
Dinh nghia
T i c h ciia vectcr a v d i so thuc k la m o t vectcr, k i hieu k. a , dxxac xac
d i n h nhir sau :

,

->

a va b la h a i vector doi nhau.

•

->

—>

_>

2)

^Z",

->

Chii y :

•

' ;• -

Neu h a i vecta a va b ciing phddng va a

vector a va k i hieu l a - a

.
•

u

V d i m o i vectd a , b va m o i so thdc k, 1, ta c6 :

1)

a

k. a cung hudrng v d i vectcr a neu k > 0 ngugc hirdng v d i vectd a
neu k < 0

a

III) Phep trii hai vectd

Vdi m6i

a)

5)

T r o n g t a m c i i a tarn g i a c

•

Dinh li
•

G la t r o n g t a m ciia t a m giac A B C o

GA + GB + GC = Q

•

G la t r o n g t a m ciia t a m gidc A B C o

OG =

OA + OB + OC


Toan

K la t r o n g t a m t a m giac Q S U nen K Q + K S + K U = 0

B a i 1. Cho tam giac A B C , goi A' la diem doi xii'ng \6i B qua A, B' la diem
doi xxjfng vdri C qua B, C la diem doi xii'ng vofi A qua C.

<^

+

sk

+ U K

dan

c:> 3 G K + ( Q P + S R
va phu y

OA = OA' + A"A

•

TiTcfng t u doi v d i OB

•

AB + BC + CA =

va

A'A = AB
Ta l a i c6 :

= OB'

(3)
Q

C

UT = - A E
2
SR

Tuongtu- OB

d

= -CA

QP

: OA = OA' + A'A = OA' + AB

CO

+ UT ) =

OC

GIAI
Ta

(2)

3 G K + ( Q K + K P ) + ( sk + K R ) + ( U K + K T ) = (D

OA + OB + O C = OA' + O B ' + O C '

•

0

=

Cong (1) va (2) ta c6 :

Chu'ng minh rang vdti mpt diem O bat ki ta c6 :

* HUdng

QK

+ B'B

= OB'

(1) (vi

A"A = AB )
QP

(2)

+ BC

(3)

OC = OC' + C'C = OC' + CA

+ UT + SR =

(3) va (4)

Cong (1), (2) va (3) ta CO :

= -EC
2

=:> 3 G K

-

2

CA + A E + EC

= 0

o

GK = 0

= 0 (4)
ci> G = K

(dpcm)

Bai 3. Cho ti? giac A B C D . Goi M, N Ian lUgft la trung diem cac canh A B ,

OA + 013 + OC = OA' + OB' + OC' +

AB + BI: + CA

C D . Chu'ng minh 2 MN = A C + BD = AD + B C
GIAI
ChiJCng minh

= OA' + OB' + O C (dpcni)
•
B a i 2. Cho luc giac A B C D E F . Goi P, Q, R, S, T, U Ian li^grt la trung diem
cac canh AB, B C , C D , D E , E F , FA.
Chu'ng minh rdng hai tam giac P R T va Q S U c6 cung trong tam.
• HUdng

dan

Goi G va K Ian lucft la t r o n g t a m A P R T va A Q S U , ta chu'ng m i n h
n

1 -

'

.

u

- -

G K = 0 bang each chu y •

GP + GR

+ GT = 0

|KQ + KS + KU =

M N = AC + B D
Ta

CO

: •

AC = A M + M N + N C

•

BD = BM + M N + N D

'

' • '
\

Cong ve, ta difcfc: AC + B1) = A M + 2 M N + N C + B M + NT)
= A M + B M + 2 M N + N C + N D (1)

0

GIAI
•

G la t r o n g t a m t a m giac P R T nen G P + G R + G T = 0
GK

<=> 3 G K

+ KP
+ (KP

+ GK
+ KR

+ KR

+ GK

+ KT) =

0

+ KT
(1)

=

0

177


GIAI

M l a t r u n g d i e m ciia A B n e n A M + B M = 0

Mk

N

•

l a t r u n g d i e m cua CD n e n N C + N D = 0

L a y d i e m O t u y y , t a c6 :
A^i

Do d6, (1) t r d t h a n h : AC + BD = 2 M N
•

+

+

+ A

\

(dpcm)
- O A j ) + ( OB2 - OA2 ) +

=

( OBj

=

( O B i + OB2 +

+ (0B„ - 0A„ )

, .

T L r o n g t u t a c 6 : AX) + BC = 2 M N

T 6 m l a i , t a c6 : 2 M N = AC +
B a i 4. Mpt gia d9 dUpc gSn v a c

B1)

tvCdng

= A D + BC

nhvC

•

hinh l a . Tam giac A B C vuong

V i n d i e m B i , B2,
B„ cung Ih n d i e m A i , A2,
hieu m p t each khac, cho n e n t a c6 :
->
OBj

c a n cf diem C. Ngifori ta treo vao diem A mpt v|it nang 5N.
Hoi C O nhang Itfc nao tac dpng vao buTc tvTdng tai hai diem B va C ?

+ 0 B „ ) - ( O A i + OA2 +

->
+ OB2 +

Tir (1) va (2)

AjBi

, An nhitog ducfc k i

^

->

+ A2B2 +

+ A^B,, = 0

+ 0 B „ = O A j + OA2 +

+ 0 A „ ) (1)

->

+ 0A„
(dpcm)

B a i 6. Cho ba diem phan bipt A, B, C .
a)

Chiing minh rSng neu c6 mpt diem I nao do va mpt so thiic t sao
>
>
>
•
cho l A = t . I B + (1 - t ) l C thi vofi mpi diem I ' ta deu c6 :
•

/ .

I-A = t.I-B + ( 1 - t ) l ' C
b)

diem A, B, C thSng hang.

Hinh l b

Hinh l a

Chu-ng to rfing l A = t . I B + ( l - t ) l C la dieu k i ^ n o^n va dii de ba
GIAI

GIAI

a)

T a i d i e m A , liTc keo F hi/dng t h i n g dijfng xuong difcJi c6 ci^cfng dp

Theo gia t h i e t : l A = t . I B + ( 1 - t ) I C , t h i v d i m o i die"m I ' , t a c6 :
IF + I'A = t.

5 N , t a c6 t h e x e m F l a t d n g cua h a i vector Fj va F2 I a n lufcft n&m
t r e n h a i dir6ng t h i n g A C v a A B .
D i thay :

f-

ir + I'B

+( i - t )

i r + rc

= t . i ' B + ( i - t ) r c + ir

| Fj I = | F | va | F2 I = I F | x/2 (do t a m gidc A B C

b)

vuong can t a i C)

I'A = t i ' B + ( i - t ) r c

N e u t a chon I ' t r u n g vdi A t h i c6 0 = t A B + (1 - t ) A C , do l a dieu
k i e n can va du de ba d i e m A , B , C t h i n g hang.

Vay : C6 m p t luc 6p vuong goc vdi biJc tirdng t a i diem C v d i ciTdng
do 5 N , v ^ m p t life keo biJc tifcrng t a i d i e m B theo hi/dng B A v d i
cudng dp "5 V2 N (Xem h i n h l b )

—>

Chiing minh rfing : A j B j + AjBg +

vi

tri

ciia

diem

G

sao

cho

GA + G B + G C + G D = 0 .

B a i 5. Cho n diem tren mSt ph^ng. B a n Minh k i h i ^ u chiing l a A i , A2, An- B a n Mai k i hipu chiing la B i , B2,

8 a i 7. Cho tii giac A B C D .
a) H a y
xac
d}nh

, B„.
+ A„'B„

b)

Chiing minh rSng vdri mpi diem O, vectof O G l a trung binh cpng
ciia bon vectcf O A , O B , OC , OD , ttfc la

= 0

OG =

- OA + OB + OC + OD (Diem G nhii the' gpi la trong tam
4

ciia tii giac A B C D ) .
17«

179

• Hudng

a i 9.

dan

Siif d u n g c o i i g thufc M A

+ MB

= 2MO

(Ola trung diem

AB).
3)

GA

Tiiih

+ GB

GC

=

+^ G D

=

OA

e)

HA + H E + H C = 2 H O

f)

Du-ofng thflng H O d i q u a t r o n g t a m G c u a t a m g i a c A B C

Vi tri cua G

OH

+ GB

= 2GI

(I la t r u n g d i e m

•

GG

+ GD

=

(J la t r u n g d i e m

2GJ

a)

AB)

ChuTng m i n h t i r a n g t\i t a c u n g co C H // B ' A

GA + GB

+ GC + G D

= 2 GI +

GA + GB

+ GC + G D

= 0

(gt)

( i ) v a (j)

GJ

nen

GI + GJ

=

Tir GA + G B

=

1

Vay

GA + GB

+ GC + G D

b)

+ GC + G D

Ta

CO

+ OA

OG

=

4

+ GO

+ OB

+ GO

+ OC

+ GO

+ OD

=

0

c)

OA + O B + OC + O D

-

A B ' C H la h i n h b i n h h a n h

AH

=2 00

OA

= OH

+ HA

= OH

= OH

- 200

=

••

=:> A H

= B'C .
B'C

=2

00

(dpcm)
-

OH

AH

- (OB

+

OC)

= 0 , t a co :
«.

GO

^

(i)

(j)

O D l a d U d n g t r u n g b i n h cua t a m g i a c B B ' C n e n

0

V a y , G l a t r u n g d i e m ciia I J
Chii-ng m i n h G O

G o i B ' l a d i e m do'i x i J n g v d i B qua O, t a co B ' C 1 B C .
V i H l a t r i r c t a m t a m g i a c A B C n e n A H 1 B C . V a y A H // B ' C

CD)

C o n g ve ciia (1) v a (2), t a co :

Ma

(difcfng

GIAI

CO :

GA

o

+ OC =

t h d n g do goi l a di^dng t h i i n g 0 - Ic c i i a t a m g i a c A B C )

•

b)

+ OB

—>

d)
(2)

GIAI
Ta

AH = 2 0D

(1)

C o n g (1) v a (2) r 6 i sU d u n g g i a t h i e t ) .

a)

C h o t a m g i a c A B C n p i t i e p t r o n g d t T o T n g t r o n (O), H l a trii'c t a m t a m
g i a c v a D l a t r u n g d i e m c a n h B C . Chiifng m i n h r S n g :

OA

+ OB

+ OC

= OH

(dpcm)

G l a t r o n g t a m t a m g i a c A B C , t a co :
HA

+ HB + HC

= 3 HG

= 3( H O

= 3 HO

B a i 8. C h o d i e m O co d i n h v a difofng t h S n g d d i q u a h a i d i e m A , B co d i n h

+ OG ) = 3 H O
+ OA

+ OB

+ 3 OG

+

OC

K e t h o p v d i k e t qua ciia cau b t a co :

Chiirng m i n h r S n g d i e m M thuQC dUcfng t h ^ n g d k h i v a c h i k h i co s*>
HA

a s a o c h o O M = a O A + (1 - a) O B
d)

Vdri d i e u k i ^ n n a o c u a a t h i M t h u Q c d o a n t h a n g A B ?
GIAI
Ta

OM

->

' ->

= uOA

+ (1 - a ) O B

< o O M - O B = a { O A - O B )
•
180

= 3Hb

+ OH

= 3 H0

- HO = 2 HO

Vi G

la t r o n g t a m t a m giac A B C

nen

3 0G

= O H , do do ba d i e m H , 0 , G t h a n g

tCr k e t

qua

(dpcm)
cau

b t a co

hang.

CO :

->
•

+ HB + HC

Vi B M = a BA

->
OM

- > - > • - >
= u( O A

o B M = a B A

a i 10. C h o t a m g i a c A B C v a d i e m O t u y y.

~ OB ) + OB _

a)

1
H a y x a c d i n h v i t r i d i e m M sao c h o O M = — 3 0 B

b)

Vdri d i e m M d a dtfgfc x a c d i n h d c a u a , t i n h A M theo A B v a

+ OC

o M e d

n e n M thuoc d o a n t h S n g A B k h i va chi k h i 0 < a

AC

^
181

:


• HUdng

a)
b)

V$y F \k dinh thu- tiT cua hinh binh hknh

C B . (xem hinh ve)

d&n

T i n h B M theo B C (can ciJ gia t h i e t )
Cho O = A t a CO k e t qua.

Ta da c6 : CD = A B ; A E

GlAl
a)

=

^

30B + OC

o

4 OB + B M

cj.

4 0 B + 4 B M = 4 0 B + BC

= SOB +

b)
<=> 4 0 M

So sanh MA + MB + MC va MD + M E + M F

=
M D + M E + M F = ( M A + AD ) + ( M B + BE ) + ( M C

OB + BC

= (MA

<=> 4 B M

Vay d i e m M a t r e n doan t h i n g BC
b)

Chon d i e m 0 = A t a di/oc : A M
• -

= BC

<:=> B M =

AE

-BC
4

•

= -

SAB + AC
4(

=

MC

+

'

Tifcfng tir : B la t r u n g d i e m D F

• 2F'C

AB;

(1)
'

: '

vi CBFA la hinh binh hanh

•

=>AE

= - A F

= : > A l a t r u n g d i e m EF.

C la t r u n g d i e m D E n e n :

;

= DE + DF

=

FD

+ F"E

2 E B = ED +

M E = MA + B C ; M F = MB + C A .

E"F

Cong ve theo vf, t a ducfc :

Chu'ng minh rfing cac diem D, E , F khong phu thupc vao vi tri ciia
diem M,
b)

= C"B

A E + A F = B1: + C B = 0

- > - > - >

MD

= BC chufng m i n h t r e n

=>

2DA

Hay xac dinh cac diem D, E , F sao cho

+ M B + M C ) + ( A D + B E + CF )

>; .

AF

B M = - B C ( h i n h ve)
4

->

+ C¥ )

Ta l a i CO : <

B a i 11. Cho tam giac A B C va diem M tuy y,
a)

va B F = CA n e n D, E, F k h o n g

phu thuoc vao v i t r i cua M

V i t r i cua M
Ta CO : Q M

= BI:

ve tren hai canh C A v^

2(D'A + ¥C + E'B)

So sanh hai tong vectcf MA + MB + MC va MD + M E + M F

o D A + F C

=

(D'E

+ ED) + (DF + FD) + (FE + EF)

+ EB = ( ) o A b + B E

= 0

'

+ CF = 0 ( 2 )

v

GIAI
a)

•

T i r ( l ) va (2)

Xac dinh diem D

Ta CO: M D = M C + A B
o

M D - MC = AB

o

a i 12.

•

Xac dinh diem E

TLforng tir, t a c6 : M E = M A + BC

o

AE =

«

182

dan

Ta CO :
.

CH 1 AB
C H // A B '

AB' ±

AB

Tuong tir A H / / C B '

(2)

TCr (1) va (2) t a co A H C B ' la h i n h b i n h h a n h

Xac diuh diem F

Tucfng t y , t a c6 : M F = M B + CA

Cho tam giac A B C npi tiep trong difoTng tron (O).

HiCdng

B"C

Vay : E la d i n h thu" t\i ciia h i n h b i n h h a n h ve t r e n h a i can''
BA va B C (xem h i n h ve)

(dpcm)

Gpi H la trUc tam tam giac A B C va B' la diem doi xrfng vdfi B qua
tam O.
->
->
->
Hay so sanh cac vector A H va B ' C , A B ' va H C .

CD = A B

Vay : D la d i n h thu" tir ciia h i n h b i n h h a n h ve t r e n h a i caul'
A B va AC (xem h i n h ve)

=> M D + M E + M F = M A + M B + M C

o

Bli' = CA

A H = B'C va A B ' = H C

• '

• ' '

183


Bai

13. C h o h a i h i n h b i n h h a n h A B C D v a A B ' C ' D ' c6 c h u n g d i n h A .

K h i do u = - 4 O M va do do

Chufng m i n h r g n g :
a)

BB' + C'C + DD' = 0

b)

H a i tarn g i a c B C D v a B ' C D ' c6 c u n g t r p n g tarn.

Chu y : Cucli chgn dicin O sao cho v = 0

'

V

= ( A B ' - A B ) + ( AC - A C ' ) + ( A D ' - A D )

A B ' f A D ' = A C ( V i A B ' C ' D ' la h i n h b i n h h a n h )

B B ' + C'C + D D ' = A C ' - A C ' - A C + AC = 0

AM

=

^ • •• •

., •

^

GIAI

,

Goi 0 , O' I a n luat la t r u n g diem ciia A D va BC, ta c6 :

= ( G B ' + B'B ) + ( G C + C C ' ) + ( G D ' + D ' D )
= ( G B ' + GC + G D ' ) + ( B'B

+ CC' +

D'b )

^ ( G B ' + GC + G D ' ) - ( B B ' + C'C + D D ' )

00'

= OA + A B + BO'

00'

= ob

Neu G la t r o n g t a m t a m giac B C D t h i GIB + G C + G D do til (*) ta cung c6 G B ' + GC + G D ' = 0

0, i u c

OA

Ma

G cung la t r o n g t a m

tam giac B ' C D ' (dpcm)

+ D C + CO'

2 0 0 ' = ( OA +

= G B ' + GC + G D ' - 0 = G B ' + GC + G D ' ) ( )

Nen

+ ob

0"b ) + ( A'B + D C ) + ( B O ' + C O ' )
=

0 (vi O la t r u n g di§m A D )

BO' + CO' =

0 (vi O' la t r u n g d i e m BC)

2 0 0 ' = A B + DC

o

00'

=

Vay t r o n g t a m hai t a m giac B C D va B ' C D ' t r i i n g nhau.
14. C h o t a m g i a c A B C v a di^ofng t h S n g d. T i m d i e m M t r e n dufong
> ^

t h A n g d sao cho vectof u

+ GC

k A B va D N = k D C .

T i m t a p hdp c a c t r u n g d i e m 1 c i i a d o a n thfing M N .

(dpcm)'

Vdi diem G bat k i ta c6 :

Bai

•

v = 0 ta chon diem O sao cho GO = —GC
4

H a i t a m g i a c B C D v a B ' C D ' c6 c u n g t r o n g t a m

G"B + GC' + 0 0

'

i 15. C h o til" g i a c A B C D . Vdri so k tuy y, l a y c a c d i e m M v a N sao c h o

AIB + A D = AC ( V i A B C D la h i n h b i n h h a n h )

b)

''

= (OA + OB + OC) + OC = 3OG + OG + GC = 40G

Vay dc

= ( A B ' + A D ' ) - A C ' - ( A B + A D ) + AC

Nen

,

G la. trpng tam tam giac ABC, ta c6 :

B B ' + CC' + D D '

Ma

u | = 40M.

Do d l i vectcf u nho n h a t k h i va chi k h i 4 0 M nho n h a t hay M la
h i n h chieu vuong goc ciia O t r e n d.

GIAI
a)

I

- >

=

- >

->

M A + M B + 2 M C c6 dp d a i n h o n h a t .
GIAI

Vdi m o i d i e m O ta c6 : u = M A + M B + 2 M C

= OA - O M + 013 ~ O M + 2 ( 0 C - O M )

- A B + DC
2

(1)

Tuang t\i : 0 va I la t r u n g diem ciia A D va M N nen ta cung c6 :
01

=

1
2

f

I

>

1

J

2

AM + DN

k.AB

Tir (1) va (2) => 0 1 = k. 0 0 '

+ k.DC

=

O

f

k . i A B + DC
2

(2)

=> I e dir6ng thSng ( 0 0 ' )

Vay k h i k thay doi, tap hop cac d i e m I la dUcfng t h i n g
M

00'.

B

= OA + OB + 2 0 C - 4 0 M
Ta chon d i e m 0 sao cho v = OA + O B + 2 OC = 0
184

185

B a i 16. C h o t a m g i a c A B C , l a y d i e m D t r e n c a n h B C s a o c h o B C = - B D .
3
Gpi I la diem xac dinh bdi 4 lA + 2 I B + 3 I C

=

Bai

A D B , B C D v a A C D , G v a G ' I a n lUfft l a t r p n g t a m c a c t a m g i a c A B C
va PQR.

0.

C h u ' n g m i n h diToTng t h ^ n g G G ' d i q u a D .

Chii-ng m i n h A , I , D nftm tr%n m p t diXdng t h ^ n g .
• HUdng

• HU&ng

dan

Dua vao cac gid t h i e t , chiJng m i n h l A va I D cung phuong (chufng

•

GIAI

Taco:

BC

3 I C -- I B

= -BD
3

<^

( ->

-> A

»

•

Ta CO : 3 D ^ = D A + D B + DC (1) v i G la t r o n g t a m t a m giac A B C

•

Tu'ong tuf ta cung c6 :

(1)
3 D G ' = DP + DQ + DR

.

Theo gia t h i e t ; 4 l A + 2 I B + 3 IC = 0

(2)

Thay (1) vao (2) ta c6 : 4 l A + 5 I D = 0

o

3DP=Db

17. C h o t a m g i a c A B C , M
• ' ->
->
MA = 3 C M
NA

= 2BN

+

va N la hai diem

dxicfc

xac

(3)

3 D Q = D"B + D"C + D D = D B + DC

(4)

d i n h bcfi

3D1I

= DA

+

DC +

m)

= DA

•

+ DC

(5)

Cpng (3), (4), (5) ta dvtac :
= 2 D A + D B + DC

Tir (1), (2) va (6), ta CO : 9 D G ' = 6 D G
->

->

Vay D G ' va D G cung phiJcfng

•

Tir M A = 3 C M

o

4 B M

.

Tir N A = 2 B N + 3 C N

«

B A - B N - 2 B N + s f s N - B^C

3B1:

V-

(6)

SC'N

dan

=

^

+ D A + DB = D A + D B

DP + DQ + DR

C h i a n g m i n h b a d i e m M , N , B th&ng h a n g .
• HU&ng

(2)

IA = - | ID

I , A , D t h a n g h a n g (dpcm)

l A va I D cung phu'cfng

Bai

1,'

GIAI

5B1)

2,'&C =

J

3IC + 2IB = SID

A p dung t i n h chat sau :

M A + M B + M C = 3 M G de chu'ng m i n h D G va D G ' cung phuang.

= 5 I D -- I B
)

dan

G la t r o n g t a m t a m giac ABC va M la d i e m tuy y, ta c6 :

minh lA = k I D )

•

C h o tur g i a c A B C D . P , Q, R l ^ n Ivitft l a t r o n g t a m c a c t a m g i a c

18.

o

BA

+ BA

(1)

-

B M

=

3 BM -

o

DG' = - DG
3

D, G, G' t h a n g h a n g (dpcm)

BC

V

o

6 B N

•

Tir (1) v^ ( 2 ) t a dirac : 4 B M = 6 B N
•

186

->

= 3B"C

~*

+ BA

(2)

«

B M = - B N
2

B M va B N cung phiictng => B , M , N t h ^ n g h a n g (dpcm)
187


TRgC

C h u v o n .lo 2

D i n h li

TOfi D O T R E N T R G C

N e u h a i d i e m A v a B t r e n t r u e x ' O x I a n lu'cft eo t o a dp l a a v a b t h i

«

t h i f c cvi

Kicn
1)

ban

AB

True

4)

Dinh nghia
True

CO

t o a dp b - a

H(? t h u - c Salof

(Chales)

l o a do ( h a y t r u e ) l a m o t d u d n g t h f l n g t r e n do d a c h o n m o t

Vcfi ba d i e m A , B , C t r e n t r u e x ' O x c6 t h i J t u t u y y , t a c6 :

d i e m O l a m goc v a m o t v e c t o u c6 do d a i b f t n g 1 ( d o n v i c h i e u d a i )
l a m vecto don v i .

•

,

^

,"> 9
2)

i >

•

T r e n t r u e x'Ox, cho vecto u .

Toan

u va V e u n g p h u o n g n e n u = a. i
•>

(a e R )

>

So a t r o n g d a n g thiJc u = a. i

>

duoc g o i l a t o a do ciia v e c t o u t r e n '

Chu y :

*

o

u v a v c6 t o a do b a n g n h a u

•

T o a do cua vecto' 0 l a 0.

•

T o a do cua A B , k i h i e u : A B (doe l a : do d a i d a i so ciia v e c t o A B )
Ta

CO

a)

T i m t p a dp x c i i a d i e m M s a c c h o M A = k M B , k

b)

T i m t p a dp t r u n g d i e m I c u a d o a n t h i i n g A B .

c)

T i m t p a dp c i i a d i e m M s a o c h o 2 M A = - 5 M B

• Hitdng

t r u e d a cho.

u va v b a n g n h a u

: AB = AB . i

n i - u - i A T D - To
Phan biet : A B va A B

T i n h t o a do ciia M A v a M B

b)

lA + IB = 0

c)

A p d u n g k e t q u a eau a v d i k =

A B l a m o t vectcf

Tim toa (JQ X ciia M

Ma

u v a V l a h a i v e c t o t r e n e u n g m o t t r u e I a n lu'cft eo t o a do a, b.

T a eo :

Vay

->

a)

u +

V

CO

t o a do = a + b

b)

u

V

CO

t o a do = a - b

c)

k u CO t o a do = k a ( k l a m o t so t h u e )

3)

T o a do ciia m o t d i e m

"

o
b)

•

M A = (a - x) i

•

M B = (b -

X)

i

(l-k)x

Toa do trung

= a - k b

C h o d i e m M a t r e n t r u e x ' O x , t a c6 : O M = m . i
T o a do m eua v e c t o O M ducfc g o i l a t o a do eiia M
OM = m. i

<=>

m l a t e a dp cvia M

( m e R)

diem

<=>a-xi=xi-b

Can

=> k M B = k ( b - x ) i

• a - kb
x = - — ^
1 - k

o

nh& : Pliaii

(k /

1)

I ciia AB
o

Dinh nghia
^

(1)

tCr (1) t a eo : a - x = k ( b - x )

I l a t r u n g d j g m ciia A B

—>

2

T h e o de b a i : M A = k M ^ B ( k ^ 1 ) '

l a m o t so' t h i f c

-->

5

GIAI

Dinh li
Neu

1.

ddn

a)

a)

AB

»

B a i 19. T r e n t r u e x ' O x , c h o h a i d i e m A v a B c 6 t p a dp I a n lUpft b S n g a v a b

*

Dinh nghia

•

AC = AB + BC

•

T o a d o c i i a vectof t r e n t r u e

Vi

,

Dinh H

o x

hict

=

l A = B I vdi

lA

= a -

Bl

= xi - h

Xj

a + b

hai cdiig t/n/'c

AC

= AB + BC

(1)

AC

=AB

(2)

+ BC

189


•

•
c)

Cong
dung
khong
Cong
hang).

thiic (1) dung trong moi tritang hap nghia Id cong thi'ic (1)
khi A, B, C d tren cung mot true (thdng hang) hoac A, B, C
thdng hang.
thijCc (2) chi dung khi A, B, C d tren cung mot true
(thdng

Chiang minh
AB

A D BC = ( d - a)(c - b ) = cd - bd - ac + ab

0

,ru

Cong ve theo ve t a dugc : A B . CD + AC . D B + A D . BC

2(a - X M ) = - 5 ( b - X M )

<:i>

= 2a + 5b o

7XM

XM

=

2a + 5b

b)

ChuTng minh

o

1

MA = - - MB
2

XM

=

u
hay

XM

=

a + c
1 +

J

a + b + c + d

=

b + d.

Tirong tir : k + 1 =

2a + 5b
— - —
7

a + b + c+ d

Do do : i + j = k + 1

B a i 20. T r e n true x'Ox, eho ba diem A, B, C c6 toa dp Ian Itf^t la a, b, c.

Ma

Tim tpa dp cua diem 1 sao cho lA + I B + I C = 0

2

k +1

o i + j
2

2

l a toa do t r u n g d i e m ciia I J

. k +1
va
l a toa do t r u n g d i e m cua K L
2

• Hiicfng ddn
Goi toa dp ciia I la x, ta c6 :
*

=

=>

Ap dung k e t qua cau a v d i k = - — , ta c6 :
5,
^ + 2
~
1 + 5
2

=0

Goi i , j , k, 1 \An liTcft la toa do cua I , J , K, L, t a c6 :

C a c h khac :
2 MA = - 5 MB

CD = ( b - a)(d - c ) = b d - be - ad + ac

AC . D B = ( c - a)(b - d ) = b c - cd - ab + ad

Tog. dQ cua M
2 M A = - 5 M B

<=>

GlAl
a)

Vay I J va K L c6 chung t r u n g d i e m (dpcm).

i A = a - x,
GlAI

Goi toa do cua I la x

•

B a i 22. T r e n true (O; i ), cho ba diem A ( - 4), B ( - 5), C(3).

Tac6:iA=a-x, IB=b-x,
Dod6:TA+!B + IC=0
,
.
< » a - x + b - x + c- x = 0

I C = c - x

Tim diem M tren true da cho sao cho MA + MB + MC = 0 . Sau
^, ^, ^ MA
.MB
do tinh =
va
MB
MC

a + b + c
o x

=

GlAl

3
B a i 21. T r e n true x'Ox cho bon diem A, B, C, D tuy y. ChiJng minh :
a)

AB.CD + AC.DB + AD.BC = 0

b)

Goi I, J , K, L Ian l\i(ft la trung diem cac canh AC, BD, AB, CD.
Chtfug minh rhng I J va K L c6 chung trung diem.

• HUdng ddn
a)

Goi toa do cua A, B, ..... l a a, b,
T i n h A B . CD =

b)

(theo a, b,

)

T i m toa do ciia I , J , K, L r o i t i m toa do t r u n g d i e m cua I J vk K L
(xem l a i cau b bai 19)

iQn

Sau d6 t i n h A B , CD ,

M A + M B + MC = 0

o

3 M O + OA + OB + OC = 0

o

O M = - (OA
3

o

O M = - (- 4 - 5 + 3) = - 2
3
MA

+ OB + OC)

o
Vay

O M = -(OA
5^'

+ OB

+

OC)

M ( - 2)

= ' O A - O M = - 4 + 2 = -2

Tuong t u t a cung c6 : M B = - 3, M C = 5
Vay

MA
MB

2

MB

_

J3

~ 3 ' MC

"

5
191


B a i 23. Cho a, b, c, d thu"

a)

la tpa dp ciia cac diem A, B, C, D tren true x'Ox

b)

MA^ + M B ^ = ( M I + I A )

Chiirng minh rang khi a + b ^ c + d thi ta ludn tim diidc diem M sao

= 2 M I % I A ^ + I B ^ + 2 M I (lA + I B ) = 2 M I % 2!A^

cho M A . M B = M C . M D

(vi l A + I B = 0 va I B ^ = lA^).

Ap dung : Tim tpa dp ciia M, neu co, biet A ( - 2), B(5), C(3), D ( - 1)
b)

K h i AB va CD co cung trung diem thi diem M of cau a co xac dinh
khong ? T a i sao ?

c)

M A ^ - M B ^ = ( M I + I A ) ^ -- ((Mi
MI +
+ IB)

= ( M I + LA)^ - ( M ! - I A ) ^ = 4IA.MI

GIAI
a)

MA.MB = MC.MD

o

(OA -

o

OM(OD + O C ~ O A - O B )

CO

O M ) (OB -

Ma

OM)

= (OC -

=

OM) (OD

-

OM)

i A . - A I =

2

nen :

MI = - I M

OC.OD-OA.OB

MA

OM (d + c - a - b) = cd - ab

+ (MI + I B )

- MB = 4

AB

(-IM)

-

2AB.IM

(dpcm).

V i a + b ^ c + d lien c + d - a - b?^0, vay : OM = — — — ~ ~ ~ r '

d + c- a - b
•
Ap dung :
Vdi a = - 2, b = 5, c = 3, d = - 1, ta thay : a + b

c + d nen diem M

d M c xac dinh va ta co :
OM =

cd - ab

_
^

d + c- a- b
b)

3.(-l) -

— i —

(-2).5

-1 + 3 + 2 -

= - 7. Vay M(- 7)

5

G i a siif A B va C D co ciing trung diem I, khi do :
OA + OB

OC + OD
-

I

^\

I = Oil

1

•

-

U K

•

(xem lai cau b bai 19)

2
2
hay a + b = c + d. Vay diem M khong xac dinh.

B a i 24. Cho A, B la hai diem tren true (O; i ) va I la trung diem ciia doan
AB. Chufng minh rang vdi moi diem M ta luon co :

a)

M A . M B = MI^ - lA^

b)

MA

+ MB

= 2 MI

c)

MA

- MB

= 2 AB.IM

+ 2 lA

GIAI
I)

•

M A . M B = (MI + I A ) (MI + I B ) '

= ( M I + I A ) ( M I - ! A ) = MI"^ - lA^

192

aHu9

(dpcm)
193


Cliii.yen d e 3

IV)

HE TRUC T 0 6 t>P OECfiC VUONG GOC

1)

T p a dp cua mpt diem
Dinh nghia

T r o n g mp(Oxy), eho d i e m M tiiy y. K h i do, toa do ciia vecto O M goi
la toa do ciia diem M , k l hieu M(x; y)

K i e n thi?c coT ban

—

I)

>

Tom tat : O M =

He true toa dp vuong g o c
T r o n g m a t phSng, cho true x'Ox c6 veeto don v i i , true y'Oy c6
vecto don v i j sao eho i 1 j .
Dinh nghia

He gom h a i true n o i t r e n goi la he true toa do Deeac vuong goc, k i
hieu mp(Oxy).
•

True x'Ox goi l a true hoanh do.

•

D i e m O goi l a goe toa do.

•

True y'Oy goi l a true t u n g do.

•

X .

>

—

>

o

i + y. j

M(x; y )

2)

Dinh li

a)

AB

b)

IAB I=

c)

D i e m chia doan t h a n g theo t i so' cho tru'dc

= (XB -

y•

X A ; yB - Y A )

^(XB

X A ) ' + (yB -

-

Y A ) '

Dinh li

Cho h a i dieni : A ( X A ; YA) va B ( X B ; ys)

;'

x^ ~ k . X y

II) T p a dp cua vectci
Dinh li

M A

= k.MB

1-k

ci.

(k ^ 1)

T r o n g mp(Oxy), chp vecto u tuy y. K h i do, eo duy n h a t m o t cap so
thuc (x; y) sao cho : u = x. i + y. j

X,

D^c bi§t, I l a t r u n g d i e m A B

Dinh nghIa

- * - > - >
Neu u = X . i + y. j

XA

=

+ XB

o
_

->
t h i cap so' x va y goi l a toa dp eiia vectcr' u

yi

XA +

=

Xt

do'i vdi mp(Oxy), k i hieu : u = (x; y)
T o m t a t : u = (x; y) o

u = X. i + y. j

B a i 2 5 . T r o n g m p ( O x y ) , v i e t t o a dp c i i a c a c vectot s a u :

>

III)
1)

u +

V

= (x + x'; y + y')

b)

u -

V

= {x - x'; y - y')

a)

b)

194

2 >

>

b = - i - 5 j ;
3

I u i = yjx^

Toa dp a = 2 i + 3 j
Toa d 6 b = - i - 5 j
3

k. u = (kx; k y )

Toa dp c = 3 i

+ y2
-->

2)

«

->
;

->

c = 3 i

->
;

•

d = - 2 j

GIAI

a)

d)

->

Tinh chat

T r o n g mp(Oxy), cho u - (x; y) va v = (x'; y')

c)

>

a = 2 i + 3 j

Pheptinh

Tpadpd

c=-b=(-;~5)
3

o

= -27

c> a = (2; 3)

c = 3. i + 0. j
o

c = 0 i* - 2 j

c=> c = (3; 0)
d = (0; - 2)

~>

Q u a n h $ giiJa u v a v
u va

->

V c u n g phiTOng

—>

—>

u

= v

o

X

c:>

x

~

= x' va y = y'

x'

=

y
y'

B a i 2 6 . V i e t vector u diidi
u

= (2; - 3) ;

d a n g u = x. i + y. j k h i b i e t t o a dp c i i a u :

u = (- 1; 4) ;

u = (2; 0) ;

J = (0; - 1) ;

u = (0; 0)

195


GIAI

u = (2; - 3)
—>

So sanh (1) va (2) t a dirge : A B = - 2 AC =i> A B va AC cung phirong

<=> u = 2 i - 3 j
—

>

—

>

—

>

u =(-l;4)

<=>u = - i + 4 j

u = (2; 0)

»

u=(0;-l)

<:5.u=0i

u = (0; 0)

C5>

=>

A, B, C t h i n g h^ng.

b)

»

{Z = 2 I

J = 2 i' + 0 j

+ j

(u =

u = oT + 0 j

Theo k e t qua cau a, ta c6 : A B = - 2 AC
Vay diem A chia doan thSng BC theo t i k = - 2

j)

•

(u = 0 )

Ta
+ b , y = a - b , z

CO

Nen :
•

•

= (1 + 0; - 2 + 3) c;.

Toa do ciia y = a Ta

CO

CB

: a = ( 1 ; - 2) va b = (0; 3)
= (- 1; - 5)

y

Ta

: a = ( 1 ; - 2)

=^ 2 a = (2; - 4)

b = (0; 3)

3 b = (0; 9)
o

=

^BC
3

= (3; 3)

^

>

X

'

Vay C chia doan t h a n g A B theo t i so k " =

—^

-3b

Nen : z = 2 a - 3 b

= (-3;-3)

=^ B A

= ( 1 ; 1)

X

Toa do cua z = 2 a
CO

BC

-2)

C chia doan t h i n g A B theo ti so k " = ?

b

N e n : y = (1 - 0; - 2 - 3) o
•

= (-2;

Ta t i m k " sao cho CA = k". CB

: a = ( 1 ; - 2) va b = (0; 3)
X

BA

Vay B chia doan t h ^ n g AC theo t i so' k ' =

Toa do ciia x = a + b
Ta

CO :

=2a-3b.

GIAI
•

B chia doan thdng A C theo ti so' k' = ?

Ta t i m k ' sao cho B A = k'. BC

B a i 27. Cho a = (1; - 2) va b = (0; 3).

Tim toa dp ciia cac vector x = a

A chia dog.n thdng BC theo ti so k = ?

3

B a i 29. Cho tam giac A B C vori A = (xi; yi), B = (X2; y2) va C = (X3; ya) trong
mp(Oxy). Tim tpa dp trpng tam G cua tam giac A B C
GIAI
G la t r o n g t a m ciia A B C , ta c6 :

z=(2-0;-4-9)

z=(2;-13)

o

o

ChuTng minh ba diem A, B, C thSng hang.

b)

Tim ti so ma diem A chia doan thSng B C , diem B chia doan thdng
A C va diem C chia doan thang A B .

=

Xn

Vay

, M la diem tiiy y.

- M A + M B + MC
3

Chon M = 0(0; 0), ta c6 : OG

B a i 28. Cho ba diem A = (- 1; 1), B = (1; 3), C = (- 2; 0) trong mp(Oxy)l
a)

MG

=

X,

+

Xo

+

=

- OA + OB +
3

6c

Xn

^

_ Yi + y2 + ys

^«

-

hay

G

x i + X2 + X3 _ y i + yg + y3

3

'

3

-3

GIAI
a)

B a i 30. Cho ba diem A = (4; 6), B = (5; 1), C = (1; - 3) trong mp(Oxy).

ChiJCng minh A, B, C thdng hang
Ta

CO

: A B = (XB

-

XA; y s - yA)

= (2; 2) = 2(

T

+ j )

AC = (xc - XA; y c - yA) = (- 1; - 1) = - (
196

T

+ j )

(1)
(2)

a)

Tinh chu vi cua tam giac A B C .

»

b)

Tim tpa dp tam dvCdng tron ngoai tiep tam giac A B C va ban kinh
dxfdng tron do.
197


a
a ii i3 3 . T r o n g m p ( O x y ) , c h o b a d i e m A ( - 2; - 1), B ( 0 ; 4), C ( 2 ; 2).

GIAI

T i m d i e m D sao cho A B C D la h i n h b i n h h a n h .
a)

2p - AB + EC

C h u v i c i i a tarn g i a c A B C :

Ma

AB = ^ „

Vay
b)

- x^f +

(YB

-

Y A ) '

BC = V ( x c - x „ f

+

(yc-YB)'

CA =

+

(YA

^(XA - X ( , f

-yc)'

+

OA

# HUcfng

= ^ l ' + (-5)'

= ^f26

= x/(-4f + ( - 4 f
=

+

92

= 4^2

ddn

•

ABCD la hinh binh hanh o

•

Goi (x; y) la toa do ciia D, tinh toa r j ciia AD va BC roi cho

=

AD = BC

2p = N/26 + 4N/2 + 3N/10

GIAI

T a r n I ( x ; y ) c i i a dUcfng t r o n n g o a i t i e p tarn g i a c A B C

Ta

CO :

(4-xf

lA^ = I B 2

<=>

<

4

IA2 = ic^

-

X

- 5Y + 13 = 0

x =

X

+ 3y - 7 = 0

5
^ = 2

X

+ (6-yf

Ban kinh R = lA

o

= (5-xf

(6-yf

Goi D(x; y), ta c6 :

+(l-yf

= (1-xf

+(-3-Yf

R = J| 4 +

•

ABCD la hinh binh hanh nen AD = BC

D a p so :

+ fe - 5 1 <=> R =
I

2J

-s/Tso
2

x +2 = 2

X

y + 1 = -2

y = -3

• HUc/ng

A, B, M thang hang o

AB va A M cung phi/cfng.
GIAI

•
•

M G Oy nen M(0; y)
DiTcfng thang (AB) qua M o M, A, B thSng hang

o

AB va AM cung phuorng, ma AB = (6; - 3) va AM = (4; y - 1) nen
4

y - 1

6

-3

c=> Y = - 1

D a p so :

M(0; - 1)

P ( 3 ; 4), Q ( - 1; - 3) t h a n g h a n g .

198

f5

D a p so :

C ( - 5 ; 5) ,

D ( - 3 ; 0).

ddn

•

Tim toa do AB va DC roi so sanh hai vector nay.

•

Tam I ciia hinh binh hanh ABCD la trung diem ciia AC (hoac ciia
BD)

'

B a i 3 5 . T r o n g m p ( O x y ) , c h o A ( - 2; 4), B ( - 4; 3), C ( 2 ; 1), D ( l ; 3)
a)

A, B , C C O t h a n g h a n g k h o n g ? T a i sao ?

b)

Chii'ng m i n h A B C D la mpt h i n h thang c a n , day B C

• HUdng

ddn

Giai giong bai 28

,

D a p so : I

B a i 3 2 . T r o n g m p ( O x y ) , t i m d i e m N t r e n t r u e h o a n h s a o c h o b a d i e m N,

* HUdng

B(-3;7)

h a n h nay.

dan

•

,

C h i i ' n g m i n h A B C D l a h i n h b i n h h a n h v a t i m t o a dp tarn h i n h b i n h

T i m d i e m M t r e n t r u e t u n g s a o c h o difcfng t h S n g ( A B ) d i q u a A .

M (0; y) e Oy

0

D(0; ~ 3)

A(-l;2)

•

=

B a i 3 4 . T r o n g m p ( O x y ) , c h o bo'n d i e m :

B a i 3 1 . T r o n g m a t p h d n g ( O x y ) , c h o h a i d i e m : A ( - 4; 1), B ( 2 ; - 2),

* HU&ng

AD = (x + 2; y + 1)
BC = ( 2 ; - 2 )

—
2

2

*

AD = BC

N -, 0
l7

ddn

a)

Xet phirong cua AB va BC ?

b)

Churng minh AD // BC va AB = DC, AD ^ BC.
199


GlAl

B a i 3 7 . T r o n g m p ( O x y ) , c h o t a m g i a c A B C : A(0, 6), B ( - 2; 2), C ( 4 ; 4)
a)

Chiifng m i n h A B C l a t a m g i a c v u o n g c a n .

b)

T i n h d i ? n tich tam giac A B C .

'i

• Hitdng dan

a) Tinh AB, BC, AC ta dugc ket qua.
b) Diing cong thufc tinh dien tich tam giac vuong.
GIAI

-4

-2

AB = 74 + 16 = 275

O

a) Ta CO :

a) Ta

-

AB = (- 2; - 1), BC = (6; - 2)
-1
nen AB va BC khong cung phuong
6
-2

CO

BC = ^36 + 4 = 2V1O
AB = AC
BC^ = AB^ + AC^

A, B, C khong th^ng hang (i)
b) ABCD ?
*

AD = (3; - 1) va B"C = (6; - 2)
3 - 1
~*
~*
Ta thay - = —- nen AD va BC cung phuang (ii)

-

T i r ( i ) v a ( i i ) ^ A D / / B C (1).
Ngoai ra, ta c6 :

=> ABC la tam giac vuong can tai A.
b) SAABC = -^AB.AC = - X 275.2V5 = 10 (dvdt).
Bai 38. Trong mp(Oxy), cho tam giac ABC, biet A(- 1; 5), B(- 3; 1) va
C(l; 4). Chiang minh ABC la tam giac vuong.
Tinh chu vi ciia tam giac ABC ?

AD = VlO , BC = 2 >/i0 , AB = VS , DC = Vs

• Hiidng

AB = DC va AD ^ BC (2)

AC = Vl6~+~4 = 2>/5

^ {.;;(}

dan

Tinh A B , AC, BC roi diing dinh l i Pitago dao.

Tir (1) va (2) ==> ABCD la hinh thang can c6 ddy BC, AD.

"''-^

Dap so': Tam giac A B C vuong tai A va c6 chu vi 2p = 3 N/S + 5
B a i 36. Trong mp(Oxy), chb ba diem A(- 2, - 5), B(2; 3), C(0; 4)
a) Chufng minh A, B, C khong 6 tren cung mpt dxicing thSng.

b) Tim diem D tren true hoanh Ox sac cho ABCD la hinh thang c6
CD va AB.

Bai 39. Trong mp(Oxy), cho A(- 2; 1), B(6; 0)
a) Tim diem M tren true tung sao cho tam giac AMB vuong tai M.
b) Chon M (cau a) c6 tung dp difotng.

ABCD CO phai la hinh thang vuong khong ?
• HUotng ddn

a) Giai giong cau a bai 32.
b) G o i D { x ; 0 ) e O x

Tim diem C sao cho AMBC la hinh chu" nhgit.
* Hiidng

JBT ^ «n6ri>J ^fi^H an fed) r") j-^^.j^ ,,,,

a) M(0; y) G Oy.

Dap so : D ( - 2; 0) va ABCD la hinh thang vuong tai B y^ C.
200

S^J;

'

Tinh A M ^ B M ^ A B ^
Diing dinh l i Pitago.

ABCD la hinh thang day AB va CD o AB va DC cung phuang.
Tinh A C ^ BC^ AB^ r6i so sdnh AC^ vdi AB^ + BC^.

\

ddn

b) Vi A M B = 90° nen A M B C la hinh chuT nhat
o

A M B C la hinh binh hanh o

M A = B"C
201


GIAI

GIAI

a)

«

M e Oy nen M(0; y), ta c6 :
AM^

•

65

A M ^ + B M ^ = AB^

<=>y^-y-12 = 0

<=> 2y2 - 2y + 41 = 65
<:i>y = 4 V y

= -3

(-2;-3)

ma
= (xc

-6;yc)

o

nen

MA

=

Xc

-

yc

=

6 = - 2
-3

4

= (3; 4)

OC

= (xc;yc)

nen <^

yc

•

Tinh M N . Q P

.

T i n h M N ^ N P ^ MP^ ^

=

yc

= 4

-4

-1 O

3

•

Chufng m i n h A B va DC bang nhau.

.

T i n h A B ^ B C ^ AC^ r o i dung d i n h l i Pitago dao.

^ M N P Q la h i n h b i n h h a n h .
r> M N 1 N P va M N = N P

a)

Tim giao diem ciia B C vofi phan giac trong cua goc A .

b)

Tim toa dp tam diiotng tron npi tiep tam giac A B C .

J * Hiicfng dan

o

B a i 41. Trong mp(Oxy), cho A ( - 4; 3), B ( - 1; 7). Co tim dtfdc diem C de
OABC la hinh vuong khong ? Giai thich ? Neu c6, hay cho biet toa
dp ciia C ? (O la goc toa dp)

Suf dung cong thuTc phan giac
DB

AB

DC

AC

D B = k.DC
Vi DB

= k
(k =

AB
AC

> 0)

va DC ngucfc hifdng nen DB = - k. DC

(D chia doan BC theo t i so - k )
b)

• Hxidng dan

Tiep tuc giong cau a doi v d i t a m giac A B D .

g

GIAI

T i n h OA, OB, A B r o i chufng m i n h t a m giac OAB vuong can t a i A
nen t 6 n t a i C sao cho OABC la h i n h vuong.
= OC ta se t i m diTofc toa do C.

=>

B a i 43. Trong mp(Oxy), cho tam giac A B C : A ( l ; 6), B(4; 0), C(9; 10).

a)

• Hiidng dan

TCr A B

3

Xc

-3

Chufng minh A B C D la hinh chff nh^t.

202

AB

OC

* HUdng ddn

BC

B a i 40. Trong mp(Oxy), cho bon diem A ( - 3; 0), B ( - 6; 2), C ( - 2; 8), D ( l ; 6).

•

B,

Chu-ng minh MNPQ la hinh vuong.

Dap so : C(4; - 3)

•

A

B a i 42. Trong mp(Oxy), cho bon diem M ( l ; 2), N(- 3; 5)* P(0; 9), Q(4; 6).

A M B C la h i n h b i n h h a n h

BC

T a m giac OAB vuong can t a i A.

Dap so': C(3; 4)

A M B = 90° nen A M B C la h i n h chiJ n h a t

=

5%/2

Toa do C

Ma

Chon M(0; 4)

MA

=

K h i OABC la h i n h vuong, ta c6 : A B =

Dap s6': M(0; 4) V M(0; ~ 3)

ci.

AB

VSO

Vay t o n t a i d i e m C de OABC la h i n h vuong.

T a m giac A M B vuong t a i M

b)

=

[OA^ + AB^ = OB^

BM'^ = 3 6 + y'^

o

: OA = 5, A B =: 5, OB

CO

OA

= 4 + ( y - l f

AB^ = 64 + 1 =

Ta

a)

Ta

CO

:

DB

_

AB

DC

"

AC

vcfi

{

A B = 79 + 36

= 3^/5 *

AC" = 764 + 16

=

4x/5
203

Vaiy

DB
DC
DB

3

DB =

4
va

B a i 4 5 . T r o n g m p ( O x y ) , c h o A ( - 5; 0), B ( - 1; 3), C ( 2 ; 7), D ( - 2; 4)

-DC

4

Chiyng m i n h A B C D la h i n h thoi. T i n h c h u v i v a d i ^ n tich ciia A B C D .

D C l a h a i vectd ngugc h i i d n g n e n D B

=

• Hudng

- - D C

3
(D c h i a d o a n E C theo ti so = - - ), do do :
4

.

Tinh AB

.

T i n h A B va B C

va

DC
^

=>
AB

=> A B C D l a h i n h b i n h h a n h .
BC

C h u vi ( A B C D ) = 4 A B

43
Xn

d&n

=

1

D i e n tich ( A B C D ) = - A C . B D

:

D a p s o ' : A B C D l a h i n h thoi c6 do d a i m 6 i c a n h = 5 v a d i e n t i c h = 7 (dvdt)
30
YD

=

1 .

D a p so : D

r43

7

^

B a i 4 6 . T r o n g m p ( O x y ) , c h o t a m g i a c A B C vdri A ( - 3; 0), B ( 2 ; 4) v a C ( l ; 5).
Chiang m i n h tam giac A B C c a n tai A. T i n h d i ^ n tich tam, giac A B C .

30

7 '

* Hudng

7;

b)

Ve p h a n giac trong c u a B , p h a n giac ii^y c ^ ^ D

=>

I l a t a r n dacfng t r o n noi t i e p t a m g i a c A B C .

T a CO :

lA

BA

ID

BD

BA

=

BD

=

-

tai I

dan

A

T i n h v a so s a n h A B v a A C .
T i m toa do t r u n g d i e m H ciia B C , t a c6 A H 1 B C n e n
SMBC=^AH.BC

3V5

v6i
D a p so : A B = A C = 4^
7

Vay
Ma

—

ID

lA

=

va

5

o

lA = - ID
5

I D ngMc

B a i 4 7 . T r o n g m p ( O x y ) , c h o h a i d i e m A(4; 1) v a B ( - 2; 5). T i m d i e m M t r e n
-*
nen I A

hudng

=

t r u e Ox s a o c h o t a m g i a c M A B c a n t a i M . T r o n g trUoTng hcfp

7
" 7 ID
o

=

=
-

4

5

{||.

Do do

nay

chuTng m i n h t a m g i a c M A B v u o n g c a n .
* Hudng

XT

SAABC = - (dvdt)
2

dan

•

M(x;

0)

•

T i n h M A ^ v a M B ^ roi cho M A = M B

D a p so : M ( -

1; 0)

;yD
yi

=

=
1

S a u k h i t i m dirac M , t i n h MA^, M B ^ v a A B ^ r o i d u n g d i n h l i P i t a g o .

5

+
B a i 4 8 . T r o n g m p ( O x y ) , c h o h a i d i e m A ( - 2; 3) v a B ( 3 ; 1). G o i (m; n) l a t o a

D a p so : 1(4; 5)

dp d i e m M .
B a i 4 4 . T r o n g m p ( O x y ) , t i m t o a dp t a m difofng t r o n n p i t i e p t a m g i a c A B C

Chiifng m i n h d i e u k i ? n c a n v a d i i de M of t r e n du'otng t r u n g t r i / c c i i a
d o a n t h a n g A B l a : 10m - 4 n + 3 = 0

b i e t A ( - 4; 1), B ( 2 ; - 2), C ( - 8; - 7)
* Hudng

• Hudng

dan

M cf t r e n du'dng t r u n g t r i / c ciia d o a n t h i n g A B

G i a i g i o n g b a i 40.
D a p s o ' : I ( - 3; 204

2)

dan

o

M A = M B v d i M A ^ =. ( m +

if

+

va MB^ =

+ (n -

1)^
205


GIAI

B a i 49. Trong mp(Oxy), cho diem M(2; 2). DrfcJng thiing (d) qua M va cAt tia

T a m giac A M N vuong t a i A <=> A M ^ + A N ^ = M N ^

Ox va tia Oy Ian Ividt tai A va B .

Xac dinh vi tri ciia (d) sao cho khoang each tiJf 0(0; 0) den (d) dai
nhat. Trong trxfdng hdp nay, tim tga dp ciia A va B ?

AM^
Taco :

A N ^ = (a - 3 f
MN^

Vay(*)

* IlUdng dan
Ve O H 1 (d), H e (d), chufng m i n h 0I^^< h^ng so.

= 16 + 4 =

sf

20

<=> (a + 1)^ + (b - 3)^ + (a - 3)^ + (b - 5)^ = 20

2a^ + 2b^ - 4a - 16b + 24 = 0

o

a^ + b'^ - 2a - 8b + 12 = 0 (dpcm).

B a i 51. Trong mp(Oxy), cho bon diem A ( - 2; 5), B(6; - 1), C ( - 1; - 2), D(5; 6).
Chu"ng minh rfing tx? giac A B C D npi tiep difofng tron c6 tam la
trung diem A B . A B C D la hinh gi ?

•

Ve O H 1 (d), H e (d)
O H = k h o a n g each ti^ 0(0; 0) den (d)

•

H = M , luc do O M 1 (d).

Vay k h o a n g each tCr 0(0; 0) den (d) dai n h a t

C a c h 1 : Chufng m i n h tam giac ABC vuong tai C va tam giac A D B
vuong tai D
(b^ng each t i n h A C ^ B C ^ A B ^

Ta CO O H < O M = 2 ^2 (hang so)
o

+ (b -

-3f

• HUdng ddn

GIAI

Vay niax(OH) = 2 ^2

+ (b

«

K h i O H dai n h a t t h i t a n i giac OAB vuong can t a i O n e n :
OA = OB = V2 .OM

= (a + i f

(*)

o

roi dung d i n h If Pitago dao)

! '

C a c h 2: T i m toa do t r u n g d i e m I ciia A B roi t i n h I A, I B , I C , I D .

Dap so : A B C D la hinh ehuT n h a t

(d) 1 O M .

K h i (d) 1 O M t h i t a m giac OAB vuong can t a i 0(0; 0)

B a i 52. Trong mp(Oxy), cho bon diem A(4; - 1), B ( - 2; 1), C(4; - 5), D(2; - 7).

(Vi O M vira la dtfdng cao vifa la duorng phan giac ve tii 0 ( 0 ; 0) ciia

a)

Tim diem I d tren true tung va each deu hai diem A va B.

t a m giac OAB).

b)

VoTi diem I tim dvtdc of cau a, chiJng minh rSng I cung each deu C
va D. Ket luan gi ve txi" giac A B D C ?

Vay OA = OB = O M V2

=4

* Hudng

Do do : A(4; 0) va B(0; 4)

a)

ddn

Gpi 1(0; y)

I each deu A va B nen l A = l b
T i n h lA^ va I B ^ theo y roi cho l A ^ = I B ^

Dap so : 1(0; - 3)
!• b)
B a i 50. Trong mp(Oxy), cho ba diem A(a; b), M(- 1; 3) va N(3; 5).
Chtfng minh rfing dieu k i ^ n can va dxi de tam giac AMN vuong tai

K h i CO toa do I , t i n h IC va I D ta c6 ket qua
So sanh bon doan l A , I B , I C , I D .

ArtS;:; f

Tu" giac A B D C ngi tiep ducfng t r o n t a m I , ban k i n h R = 2^5

A la : a^ + b^ - 2a - 8b + 12 = 0
• Hudng

T a m giac A M N vuong t a i A o
206

B a i 53. Trong mp(Oxy), cho ba diem A ( - 4; 6), B ( - 2; - 1), C(3; 4).

dan
A M ^ + AN'-^ = M N ^

Tim diem M tren true hoanh sao cho | MA + MB + MC I ngfin nhfi't.
207

• HUdng

dan

Sii'dung cong thufc

MA + MB + MC = 3 M G

(G la t r o n g tarn t a m

Chuyen de 4

T i s6

giac ABC)

K i e n thi?c c«r ban

GIAI

•

Ta CO M A + M B + M C = 3 M G vdi G la t r o n g t a m t a m giac ABC
3

G

=

I)

Ti so li/dng giac cua goc a e [0°; 180°]

1)

Mcf d a u

G ( - 1; 3)

+ yfi + yc

LCrONG GIfiC

T a m giac A B C vuong t a i A, t a c6 :

^ 3

B
Vay d =

I

M A + M B + MC

I =3I

MG

I=

•

3MG

•

2)

sinB =
tgB =

AC
BC
AC
AB

•

cosB =

•

cotgC =

AB
BC
AC
AB

T i so' Irfgfng g i a c c t i a m p t goc

T r o n g he true toa do Oxy, ve niira dirdng t r 6 n don \n

kinh

R = 1), nufa ducrng t r o n nay cdt true x'Ox t a i A va A', c^t t i a Oy t a i
C v d i A ' ( - 1; 0), A ( l ; 0), C(0; 1)
•

d ngan n h a t o M G ngSn n h a t ci> M G 1 Ox.
Luc do M ( - 1; 0) va min(d) = 3 M G = 9

•

Cach khac
Goi M(x; 0), t i m toa do ciia M A , M B , M C
vectcf
=

MA

= M A + M B + MC
+ (x + i f

> 9

o x

r o i suy ra toa do cu;i

u| = ^ ( - 3 - 3 x f

+ 81

hay
a)

= - l

Dinh nghia

T r e n nufa difdng t r o n don v i (.ban k i n h R = 1) lay d i e m M ( x ; y).
Dat a = A O M

B a i 54.

T r o n g m p ( O x y ) , c h o b a d i e m A ( - 4; 2), B ( - 7; 1), C ( - 1; 0).
T i m d i e m M t r e n t r u e t u n g sao c h o vectof
C O dQ d a i nhft n h a t .

* Hitdng

v

=

MA

+ MB +

•

MC

•

D a p so : M(0; 1) va m i n | v | = 12

Tung dp y eiia M goi la sin ciia goc a, k i hieu sina va t a e6 sina = .y.
H o a n h do x cua M goi la cosin cua goc a, k i hieu cosa va t a c6
cosa = X .

d&n

G i a i giong bai 53.

•
•

, y
y
T i so — (x 9t 0) goi la t a n g ciia goc a, k i hieu tga va ta c6 tga = — •
X

•

X

T i so' — (y # 0) goi la cotang ciia goc a, k i hieu cotga va ta cd
y
"
.

•
toga =

208

(0*^ < a < 180°), t a eo :

-.
y
209


b)

T i so Ivtifng giac ciia mOt s6' g6c cAn nh<3r
Goc

0°

sin

0

60°

90°

1

72

s

1

2

2

V3

V2

1

2

2

2

V

V3

0

tg

45°

2

n
1

cos

30°

3

^/3

cotg

1

73

c)

D a u ciia cac ti so Itfdng giac

•

s i n a > 0 v 6 i m o i goc a

150°

72

1

2

2

1

72

73

2

2

2

-73

- 1

0

3

135"

2

0

V3

1

120°

73
3

- 1

73
3

3)

L i e n h ^ giu^a ti so' lifofng g i a c c i i a h a i goc bik n h a u

a)

dSO" - a) va a Id hai goc bu

180°

nhau

Dinh li
H a i goc bu nhau c6 sin hhng nhau, con cosin t h i d o i nhau, n g h i a 1^

0
•

sin(180° - a)

sina

•

cos(180° - a) =

-cosa

t g ( l 8 0 ° - a ) = - tga

Do do

c o t g ( l 8 0 ° - a ) = - cotga

- 1
b)

(90 - a) va a la hai goc

pht/inhau

Dinh l i

0

,

H a i goc phu nhau t h i s i n ciia goc nay b k n g cosin cua goc k i a va
ngucfc l a i , nghia la :

-73

sin(90

- a) = cosa

cos(90° - a) = sina

e [0°; 180°]

Do do

t g ( 9 0 ° - d ) = cotga
cotg(90° - a ) = t g a

loan

V d i 0 < a < 90° ta c6 0 < cosa < 1
V d i 90° < a < 180° ta c6 - 1 < cosa < 0
•

t g a va cotga (neu khac 0) deu cung dau vdi cosa.

II) Cac he thi/c glQa cac ti so lUting glac
1)

Cac

thiifc cot b a n

Dinh l i : V d i m o i goc a, t a deu c6 :
a)

.
sina
N e u cosa 5^ 0 t n i t g a =
cosa

b)

N e u s i n a * 0 t h i cotga = s—
ina

c)

sin^a + cos^a = 1

-.-r^'

2)

C O S cx

Cac

thufc k h a c

B a i 55. Biet cosa = - , tinh P = Ssin^a + 4cosV
2
• HUcfng dan

•

B i e t cosa, dung he thiJc : cos^a + sin^a = 1 de c6 sina.

•

B i e t sina, v i n diing he thuTc do de biet cosa.
GIAI
Ta CO : cos^a + sin^a = 1 ( d i n h l i )
1
2
1
cos a - — => cos a = — n e n
2
4
1
. 2
1
.
2
,
1

— + sin a = 1 o sin a = 1
=
4
4
3
Do do : P = 3sin^a + 4cos^a - 3.
Mk

:
3

—
4
+ 4.

1

13
4

Dinh l i :
a)
b)
210

N e u cosa ^ 0, ta c6 : 1 + tg^a =

—
cos a
1

N e u s i n a ?^ 0, t a c6 : 1 + cotg a =
sin^ a
211


Jai 57. Chiang minh h^ng d^ng thu"c : (sinx + cosx)^ = 1 + 2sinx.cosx

B a i 56.
a)

HUitng dan

Cho goc nhon P va sinP = — . Tinh cosp va tgP ?
4

b)

Cho goc a va cosa = - — . Tinh sina, tga va cotga.
3

c)

Cho tgx = 2\l2 . T i n h sinx v a cosx

M a sinP = — n e n
4

+ cos% = 1 c=> cos^p = 1 - — = — (i)
^

16

•

(dinh l i ) = ^
V15

1

8

,9

9

(dpcm)

(sinx + cosx)^ = 1 + 2sinx.cosx.

Vl5
B a i 58. Chu'ng minh hSng d^ng thu^c : (sinx - cosx)^ = 1 - 2sinx.cosx

Tuong tir bai 57

=
^yl5
1\

f

3

GIAI
2

+ sin^a = 1

nia sina > 0 nen sina =

CO

tga =

sina

2V2

2

Ta CO : (sinx - cosx) = (sinx)

2V2

2

+ (cosx) - 2sinx.cosx

- sin^ x + cos^ X - 2blxix.cosx
=
1.

- 2sinx.cosx

cosa

3

f 1^

(dpcm)

B a i 59. Chufng minh h&ng dSng thii^c : sin^x + cos^x = 1 - 2sin^x.cos^x
• 'lUc/ng dan

T i n h cotga

T a CO tgx = 2J2

f M . 2V2
. 3,

ung hai cong thiJc :

cos^ x
sin

X

cos X
Ta CO —V-

COS^ X

3

1

•

A p dung cong thufc : a^ + b^ = (a + b)^ - 2ab

2V2

GIAI

> 0 n e n sinx > 0 va cosx > 0
1

A p dung cong thuTc a^ + b^ = (a + b)^ - 2ab, t a c6 :

= l + tg"x

sin'^x + cos'^x = (sin^ x + cos^ x j

= tgx

= 1 + tg\ 1 + (2 V 2 )^ = 9 o

cos^x = -

(i)

9

V i t g x > 0 n e n cosx > 0, do d6 tCr (i) t a c6 cosx = - .
o
Ta CO 7-^— = t g x o sinx = cosx.tgx = - x 2\f2
cosx
3
212

+ 2sinx.cosx

T i n h tga

cosa
sina
c)

2

* HUcfng ddn

Tu cos^a + sin^a = 1 ( d i n h l i ) , t a c6 :

Ta
•

2

Vay

16

T i n h tgP
Ta c6 : tgp = ^
cosp

b)

s i n X + cos x = 1

1

V i p Ik goc n h p n n e n cosP > 0, do d6 til (i) t a c6 : cosp =
•

•

± 2ab

Ta CO (sinx + cosx)^ = sin^ x + cos^ x + 2sinx.cosx

T i n h cosP
Ta CO : cos^p + sin^p = 1 ( d i n h l i )
1

Sir dung (a ± b)^ = a^ +

GIAI

GIAI

a)

•

=

2V2

3

t

Vay

- 2 sin^x.cos^x

1^

- 2 sin^x.cos^x

1

- 2 sin^x.cos\

sin'^x + cos''x = 1 - 2 sin^x.cos^x

B a i 60. Chu'ng minh hdng dSng thtfc :
sinx.cosxd + tgx)(l + cotgx) = 1 + 2sinx.cosx
213

' a i 63. DoTn g i a n b i e u thiiTc : C = s i n a ^ l + t g ^ v d i 9 0 ° < a < 1 8 0 ° .

• Hitdng d&n

•

HUdng d&n

Khai trien (1 + tgx)(l + cotgx) roi sau do tinh va riit gon ve tra
, ,
sin X ,
,
cos X ,
(dung cong thUc tgx =
va cotgx =
)
cos X
sm X
GIAI
1 + sinx
cosx

Ta CO : (1 + tgxXl + cotgx) =

cos x

1 + cosx
sinx )
+

sui X

+

(

Do do : sinx.cosxd + tgx)(l + cotgx) = sinx.cosx 2 +

Siir dung cong thiJc 1 + tg^a =

.

VA^ =

•

90" < a < 180°

— ^
cos a

IAI

cosa < 0
GIAI

sin X cos x
.
sinx

cosx
cosx sinx
= 1+
+ 1
+
cos
X
sin
X
sin X
cos X
cos^ X + sin^ X
= 2 +
=2+
sinx.cosx
sin X. cosx
= 1+

-

1
sin x. cos

X

•

Ta CO C = sina v/l + tg^a - sma.
si

•

Vi 90° < a < 180° nen cosa < 0
.
„
Vay C =

sina
-cosa

cos^ a

sina
cosa

=> | cosa I = -cosa

= - tga

Bai 64. Tinh : 008^12" + cos^TS'* + cos^l° + c o s W
• HUdng

ddn

Su dung dinh h've ti so' lucfng giac ciia hai g6c phu nhau.

= 2sinx.cosx + 1 (dpcni)

Cong thiJc cos^a + sin"a = 1
Bai 61. Dan gian bleu thuTc : A = cosy + siny.tgy

GIAI

•¥ HU&ng ddn

Vi 12° + 78° = 90°, nghia la 12° va 78° la hai goc phu nhau.

^
, •
. ,
sin y ,
. 2
Dung hai cong thuc tgy =
va cos y + sm y = 1
cosy
GIAI
A = cosy + siny.tgy = cosy + siny.
= cosy +

* HUdng

cos^78° •= sin2l2°
cos^l2° + cos-78° = cos-12° + sinh2° = 1

siny
cosy

Tuong tu : 1° va 89° la hai goc phu nhau nen cos98° = sinl°

sin"' y

cos" y + sin" y

1

cosy

cosy

cosy

Bai 62. Dotn gian bieu thtfc : B =

Do do : cos78° = sinl2°

+ cosb.^1 - cosb

dan

cos^l° + cos^89° = cos^l° + sin^l° = 1
Tom lai cosh2° + cos^78° + cos^l° + cos'^89° = 2
Bai 65. Tinh : T = sin^3» + sin^l5° + sin^TS" + sin's?"
* HUdng

Tirang t\l bai 64

S\i dung cong ihufc \/a.v^ = \/ab (a, b > 0) va chii y 1 - cos% = sin"!

GIAI
B = x/l + cosb.^l - cosb = 7(1 + cosb).(l - cosb)

= sjl - cos" b = \/sin^ b = ! sinb I
Vi sinb > 0 nen j sinb! = sinb, do do : B = sinb

214

dan

Ta CO T = (sin^3° + sin^87°) + (sin^l5° + sin^75°)
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GIAI
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Ta CO sin^3° + sin^87° = sin^s" + cos^3° = 1

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Tuong tiT sin"15° + sin-75° = sin^l5° + cos^l5° = 1
Vay T = sin^3° + sin^l5° + sin^75° + sin^87° = 2