Ôn Tập Toán 10 Phần Đại Số

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n tp ton 10 phn i s (lp 10C3)

GV son: PhmTun Khng Trang

I. Phn Hm S

Bi 1:Biu din cc tp hp sau trn trc sri cho kt qu:

1) 3;3 ( 2;1)

2) ;0 3;

3) 1;3 0;6

4) 3;3 \ 1;5

5) 3;2 1;3

6) \ 0;3R Bi 2:Tm tp xc nh ca hm ssau:

1)

2)2

7

2 3

xy

x x

3)

4)

2 3

2 3 2

xy

x x

5) 2 2 1y x x

6)1

3 21

y xx

7) 3

41

y xx

8)

1, x 3

2 1

4 , x< 3

xy

x

Bi 3:Xc nh tnh chn lca cc hm ssau:

1) 4 23 2y x x

2)1

y xx

3) 33 4y x x

4) 3y x

5)1

2

xy

x

6) 2

3y x

7)3

1

xy

x

8) 1 1y x x

9) 2 23 3

x xyx x

Bi 4:Tm gi trca hm s:3 8, x < 2

( )2 1, x 2

xy f x

x

ti cc im x = -3 ; x = -2 ; x = 0 ; x =

Bi 5:Xc nh hm sy = ax + b bit ng thng i qua hai im M(-1;3), N(1;2)Bi 6:Xc nh hm sy = ax + b trong cc trng hp sau:a) ng thng trn ct ng thng y = 2x + 5 ti im c honh bng -2 v ct ng thngy = -3x + 4 ti im c tung bng -2.

b) ng thng trn song song vi ng thng 12

y x v i qua giao im ca hai ng thng

11, y 3 5

2y x x

Bi 7:Xt sbin thin v vthca hm s: 1 2y x

Bi 8:Cho hm s: y = x + 2a) Xt chiu bin thin v vthhm s: y = x + 2.

b) Tm giao im ca thhm sy = x + 2 v thhm sy = -3x + 1

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c) Da vo thtrn vthhm strn hy vthhm sy = 2x

Bi 9:Xc nh hm s: y = 2x2+ bx + c bit rng thca na) C trc i xng l ng thng x = 1 v ct trc tung ti im M(0;4)

b) C nh l I(-1;-2)c) i qua hai im A(0;-1) v B(4;0)d) C honh nh l 2 v i qua im N(1;-2)Bi 10:Xc nh hm s: y = ax24x + c, bit rng thca na) i qua hai im A(1;-2), B(2;3)

b) C nh I(-2;-1)c) C honh nh l -3 v i qua im P(-2;1)d) C trc i xng l ng thng x = 2 v ct trc honh ti im Q(3;0)Bi 11:Cho hm s: y = -x2+4x -3(P)a) Xt sbin thin v vthhm s

b) Tm giao im ca (P) vi ng thng (d): y = -x + 1

c) Da vo (P) hy vthhm sy =

2

4 3x x d) Bin lun theo m snghim ca phng trnh: -x2+4x -4m = 0Bi 12:Cho hm sy = x2-2x +1 (P).a) Hy xt chiu bin thin v vth(P) ca hm s

b) Da vo (P) vtrn hy vthca hm s: 2 2 1y x x

Bi 13:Hy vthca cc hm ssau:

a) 322 xxy , b) 2 4 2y x x , c) 2 1x y = x

II.Phng trnhBi 1:Gii cc phng trnh sau:a) 2x -3x + 1 =0

b) x4 + 5x2+ 4 =0c)

d)2 1 2

3 3

x x

x x

e)

f)3 1

3

3

xx

x

g)1 2

11 2x x

k)1 3 5

2 2 2

x x

x x

h)2 2

4 5 3 3( )

3 5 3 2

x xt x

x x x x x

k) 2 2(3 5) (3 2)x x

Bi 2:Cho phng trnh: a) Gii phng trnh vi m = 3.

b)Tm m phng trnh c hai nghim tri.c) Gii v bin lun phng trnh trn theo m.Bi 3:Tm m phng trnh sau c hai nghim x1,x2thomn:a) c

b) ( ) c

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c) c 5

9

d) c . n: a) m = 0;1 b) m =-3 c) 50/57 d) -13Bi 4:Cho phng trnh () Tm m phng trnh c hai nghim phn bit v tch ca hai nghim bng 8.Bi 5:Cho phng trnh ( ) c hai nghim

.Tm m biu thc:

P = t gi trnhnht.Bi 6:Cho phng trnh ( ) c hai nghim . Tm m biu thc:P =

t gi trln nht.

. n: m = 27/8.Bi 7:Cho phng trnh: x2- 2(m- 1)x + m2 - 3m = 0

a) Gii phng trnh vi m = 2

b) Tm m phng trnh c mt nghim x = - 2. Tm nghim cn li

c) Tm m phng trnh c hai nghim phn bit . Khi tm h

thc lin h gia x1v x2khng ph thuc m .

d)Tm m phng trnh c hiu hai nghim bng 2 .e) Tm m phng trnh c hai nghim x1v x2tho mn: x1

2+ x22

= 8

f) Tm gi tr nh nht ca A = x12+ x2

2

Bi 8:

Tm m phng trnh: c ba nghim x1, x2,

x3, tho mn2 2 2

1 2 3x x x nh nht.

Bi 9: Gii cc phng trnh sau:a) 2 4 1x x

b) 4 1 2 5x x

c) 01152 xx

d) 243 xx

e) 23 5 2 3x x x

f)

2 1

2

xx

x

g) 5x+2 + 3x- 4 = 4x+5

h) 2 2 2 1 5x x

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Bi 10:Gii cc phng trnh sau:

a) 25 x x 1

b) 2 9 1 2x x x

c) 2 2 4 2x x x

d) 3x +1 x 4 1

e) 9 5 2x x 4

f) 1 x 6- x -5- 2x

g) 2 24 2 8 12 6x x x x

h) 2x + 9- x = -x +9x +9

Bi11: Gii cc phng trnh sau:

a) 225 1x x

b) 916- x x 7

c) 5x+ 7- 3x+1= x+ 3

d) 24 1 3 5 2 6x x x x

e) 22 4 6 11x x x x

f) 11 11 4x x x x

g) 2 x 2 2 x 1 x 1 4

k) 22 7 2 1 8 7 1x x x x x

h)

2 2

2x + 8x +6 + x -1 = 2(x +1)


a) 2 2x -3x +3+ x -3x +6 =3

b) 2 2 2x +x +7 + x +x +2 = 3x +3x +19

c) 22x +3 + x +1=3x -16+ 2 2x +5x +3

d) 22 1 3 1 0x x x

III. Hphng trnh.Bi 1:Gii cc hphng trnh sau:

a)3 4 2

5 3 4

x y

x y

b)4 5 3

7 3 8

x y

x y

c)

2 3 2 4

4 9 2

8 4

x y z

y z

y z

d)3 2 2

5 3 2 10

2 2 3 9

x y z

x y z

x y z

e)

6

7

11

x y z

xy yz xz

xy yz zx

f)

2 2 18

. 1 . . 1 72

x x y y

x x y y

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Bi 2: Gii cc hphng trnh sau:

a)2 2

2 7 0

2 2 4 0

x y

y x x y

b)

2 2

2 2 2 1 0

3 32 5 0

x y x y

x y

c)

2 2x +y +xy =7

x+y+xy=5

d)

2 2x + y + xy = 4

x + y + x y = 2

e)

3 3 3 3x +y +x y =17

x+y+xy=5

f)

4 4 2 2

2 2

x +y +x y = 21

x +y + xy =7

Bi 3: Gii cc hphng trnh sau:

a)

2

2

x =3x-4y

y =3y-4x

b)

2

2

x =3x+2y

y =3y+2x

c)

2 2

2 2

x - 2y = 2x + y

y - 2x = 2y + x

d)

1 32x + =

y x

1 32y + =

x y


a) 2x - 2 x + 2 = -2x + 2 +2 x - 4

b) x + 1+ 4- x + x +1 4 - x =5

c) 3 312 14 2x x

d) 3 3 32 3 2 1x x x

e) 4 456 41 5x x

Bi 5:Gii cc pt trnh sau:

a) 3 x-2 + x+1= 3

b) (A-2009 ): 2 3 3x -2 +3 6-5x -8 = 0

c) x+ 2 217 x x 17 x 9

d) 2 2 2 23 7 3 2 3 5 1 3 4x x x x x x x (Lin hp: x = 2)

e) 3x +1 x 4 1( Lin hp: x =5)

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f) 3 3 2x -1 x ( Lin hp: x = 2)

g)1

4 1 3 2 ( 3)5

x x x ( lin hp: x= 2)

h)

3 2 2 2 6x x x ( Lin hp: x = 3; x =

11 3 5

2

)

k)23 1 6 3 14 8 0x x x x

Bi6: Gii h phng trnh:

/a)

2

2 2 9

4 6

x x x y

x x y

b)

2

2

1 4

1 2

x y x y y

x x y y

c)2 2

2 2

3 4 1

3 2 9 8 3

x y x y

x y x y

d)3 3 3

2 2

1 19

6

x y x

y xy x

f) 2 2 21 7

1 13

xy x y

x y xy y

g)

2

2

1 3 0

51 0

x x y

x yx

...

h)

2 3 2

4 2

5x y x y xy xy

4

5x y xy 1 2x

4

g)

3 3 3

2 2

y x 9 x

x y y 6x

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e)

2 2 2 2

11 5

11 49

x yxy

x yx y

IV. Bt phng trnh, hbt phng trnhBi 1:Tm iu kin ri suy ra tp xc nh ca cc btphng trnh sau:

a)

2

1 12

x 3x 1

b) x 5 1 3 2x c)

2

2

x 12x 3 0

x 2

d)

2

x 1x 1

x 2

e)x 1

x 3 x 3

f)1

3 x 1 2xx 2

g)

x 1 1 1

x 2 x 3 x 4x 1

h) 232

1 x2x 1

x 3x 2

Bi 2:Gii cc bt phng trnh sau:

a) x 3

2 x 1 x 33

b) x 7 x 6 x 1 x 2 x

c) x 2 . x 3. x 4 0

d) 2

x 1 x 2 0

e) x 2 x 3 x 4 0

f) 1 x 3 2 1 x 5 1 x 3

g) x 4 x 5

2x 5

Bi 3:Gii cc hbt phng trnh sau v biu din tp nghim trn trc s:

a)

33x x 2

5

6x 32x 1

2

b)

4x 5x 3

6

7x 42x 3

3

c)

3 2x 732x

5 3

5 3x 11

x2 2

d)

3x 1 3 x x 1 2x 1

2 3 4 3

2x 1 43 x

5 3

Bi 4:Gii v bin lun btphng trnhsau theo m:

a) 2mx m 2x 4

b) x 1 m x 2

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Bi 5:Tm m hbt phng trnh sau c nghim:2x 4m 2mx 1

3x 2 2x 1

HD: Nu m x 1 2m => hc nghim khi -3 < 1+ 2m hay m > -2. Kt hp vi km -2 < m 0.x 0 lun ng=> hlun c nghim x > - 3Nu m > 1/2 th (1) => x 1 2m => hlun c nghim x 1 2m .Vy hc nghim khi m > -2

Bi 6:Tm m hbt phng trnh sau v nghim:2mx 9 3x m

4x 1 x 6

.n: 2 m 3

Bi 7: Xt du cc biu thc sau bng cch lp bng:

a) f (x) 3x 1 x 2 x 4

b)

2x 1

f(x) x 1 x 3

c)1 1

f(x)3 x 3 x

d)2

2

x 6x 8f(x)

x 8x 9

e)

2x 1f(x)

x 1 x 3

f)2

4 2

x 4x 4f(x)

x 2x


a) 1 2x x 2 x 4 0

b) 2 13 x

c) 4x 1 33x 1

d)2

2

x x 31

x 4

e)1 1 1

x 1 x 2 x 2

f) x 2 2x 1 x 1

A. BT PHNG TRNH CHA N TRONG DU TRTUYT I

L thuyt:Dng cbn:

2 2

f (x) a1) f (x) a a f (x) a

f (x) a

f (x) 0

f (x) af (x) a2) f (x) a hoc f (x) af (x) a f (x) 0

f (x) a

3) f (x) g(x) f (x) g (x)


a) 3x 5 2 e)2 x

2x 1

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b) 5 8x 11

c) x 2 2x 3

d) x 1 x x 2

f)x 1

2x 1

h)x 1

1x 2

Bi 10: Gii cc bt phng trnh sau:

a) 2 x 1 x 4 b) x 1 2 x 1

c)x 3 x

1x 2

d)2

3x1

x 4

e)2

x 6 x 5x 9 f) 2x 2x 3 3x 3

g) 2x x 5 0

h) 2x 6x 7 x 6

k)2

2

x 3x 13

x x 1

* Bi luyn tp:Gii cc bt phng trnh sau:

2

2

2

2 2

2

2 2

1) 6 ( 6 1 7)

2) 5 6 ( 1 2 3 6)

3) 5 4 2 ( 2 2 4)

1 14) 3 2 1 ( )

4 2

5) 5 9 6 (1 3)

6) 2 4 0 ( 2 1)1

7) 1 2 ( )2

8) 1 2 3 ( 0 2)

29) 3 5 3 ( )

3

x x x x

x x x x

x x x x x

x x x x x

x x x x

x x x x x

x x x

x x x x x

x x x x

2

2

2

1) 2 4 2 , ( 3 5)

4 22) 1 , ( )

2 5

2 53) 1 0 (3 2)

3

2 104) 3 (3 )

5 6 3

26) 2 (0 1)

2 3 3 1 1 39) 1 ( )

1 4 2 4 2

x x x x x

x xx

x x

xx

x

xx

x x

x xx

x

xx x

x

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B. BT PHNG TRNH CHA N TRONG DU CN

L thuyt:Dng cbn:

2

2

f (x) 0 f (x) 0f (x) 0

1) f (x) g(x) g(x) 0 ; 2) f (x) g(x) g(x) 0f(x) g(x)

f(x) g(x) f(x) g (x)

g(x) 0g(x) 0

f (x) 0f (x) 0

3) f (x) g(x) g(x) 0g(x) 0

f (x) 0

f(x) g (x)

2f(x) g (x)

B

11: Gii cc bt phng trnh sau:

a) 2x 3x 5 x 1

b) 2x 2x 1 x

c) 2 4 5 2 3 x x x

d) 22x 6x 1 x 2 0 e) 3x < 2x

f) 7 1 3 18 2 7 x x x g) 2 21 1 3 x x

Bi 12: Gii cc bt phng trnh sau:

a) 262 xxx ( 3x )

b) 1)1(2 2 xx ( 311 xx )

c) xxx 122 ( 4x )

e) xxx 2652 2 ( 110 xx )

f) 12411 xxx ( 54 x )

g) 1553 xx ( 4x )

h) xxx 12 (3

323 x )

k)3

73

3

)16(2 2

x

xx

x

x

* Bi luyn tp:Gii cc bt phng trnh sau:

1x

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2

2

1) 7 6 3 2 (1 6)

12) 4 1 ( 4 0 )

6

73) ( 1)(4 ) 2 ( 1 )

2

x x x x

x x x x x

x x x x x

4) xxx 31415 ( 1

4x )

5) 5 1 1 2 4 ( 10 2) x x x x x

6) 2 35 4 3 ( 3 4 )3

x x x x x

7)2 22 5 4 2 4 3

( 1; 1 2 6; 1 2 6; 1)

x x x x

x x x x

8) 2 22 4 3 3 2 1 ( 3; 1) x x x x x x

9) xxxx 271105 22 ( 13 xx )

10) 2855)4)(1( 2 xxxx (9< x< 4)

11)2 2 23 2 4 3 2 5 4x x x x x x

( 4x V x =1)12) 2 2x 2 x 3x 4 x 4 x 2 4 x 8

13)21 1 4 2 1 1

3 ( ; ;04 2 2

xx x x

x

15)23x 16x 5 1

2 x 1 3 x 5x 1 3

16) 1 8x 3

4 x4x

Bi7Gii cc BPT sau:

a) 25

2 3 5 1 32

x x x x

b) 2 2 1 2x x x x

c) 2 23 3

1 3 1 1 ; 12 2

x x x x

d) x + 2 - x - 6 > 2 ( x 6 7)

e) 2 3 42x +1 x x - 3 x

f) x +3 - x -1 x - 2 2 213

x

; x 1 2 x 2 5x 1 2 x 3

g) (x+2) x x x 2 23 4 4; ( 2;4 8x x ) HD : x 2 x 2 0 ...

h)21 1 4 1 1

3 ; 02 2

xx x

x

HD: lin hp: 22

04

3; 04 3. 1 1 41 1 4

xx

xx xx

k) 2 2 23 2 4 3 2 5 4 4; 1x x x x x x x x

m) 5 2 3 3 0 4; 1x x x x x x

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n) 2 21

4 3 2 3 1 1 1;2

x x x x x x x

HD: phn tch thnh tch

i) 224 4 2 2 2 3 2 3 x x x x x x

HD:

2 2 2 3 2 24 4 4 2 0 dat: t = 4 >0 ta co: -t -t +20 t>x x x x x x x x t t t

j) 2 2x 25

x 4 8 x x HD:dat t= x 4,t 0 t x 4 taco: t 2 2 4 t4 4

2 32t t 6 0 t2

...

p) (A-2010):

x x

x x21

1 2 1

. HD: ta c:

x x x

2

2 1 3 32 1 22 2 2

MS 0

Do vi x 0 BPT tng ng vi: x x x x22 1 1 . Thy x = 0 khng l nghim

ca BPT

Nn x>0. Chia c2 vcho x >0 ta c: x xx x

1 12 1 1

.

t t = x x txx2

1 12 ta c:

tt t t

t t2

2

12 1 1 1

2 1 0

x x x x xx

1 1 5 3 51 1 0

2 2

C). BI TP NGH

Bi 5:Gii v bin lun cc bt phng trnh sau:

1) mx2 x

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2) 3x2 2 < xm

3) mx m2x > m3x

III. PHNG PHP HM S

Phng php ny da vo vic kho st mt vi tnh cht c bit no ca hm sdn n kt lun nghim cho phng trnh, bt phng trnh ang xt.

V d: Gii bt phng trnh: 9 2 4 5x x .

Gii:

Xt hm s 9 2 4y x x , ta thy ngay hm sny ng bin trn tp xc nh 2x .Tc f(0) = 5 do :

+ Vi x > 0 th f(x) > f(0) = 5 nn x > 0 l nghim.

+ Vi 2 0 ( ) (0) 5x f x f nn 2 0x khng l nghim.Tm li: x>0 l nghim.

IV. P DNG BT NG THC

1). MT SV D

V d1:Tm gi trln nht ca hm s 2 4y x x v p dng gii phng trnh:22 4 6 11x x x x .

Gii:p dng bt ng thc : 2 2 22( ) ( )a b a b .ta c:

2

2( 2 4 ) 2 4 2x x x x y . Do y ln nht bng 2 khi v chkhi:

2 4 3x x x .Mt khc 2 26 11 ( 3) 2 2.x x x x nn:

2

2

2 4 22 4 6 11 3

6 11 2

x xx x x x x

x x

V d2:Gii phng trnh

x3 +x

1= 4 8 x (1)

Gii.

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MX: x > 0

C4

x

1x3

=8

x

1

x

1xxxxxx

8 x (2) x > 0 (BT Csi)

Vy (1) du = (2) xy ra x =x

1 x = 1.

V d3:Gii phng trnh

2x + x4 = 2x 6x + 11. (1)

Gii.

* Cch 1

2)1(VT ( 21 + 21 )(x2 + 4x) = 4. (BT Bunhiacopxki)

VT 2.

VP(1) = 2)3x( + 2 2.

Vy (1)

2)1(VP

2)1(VT

03x

1

x4

1

2x

x = 3.

* Cch 2

t A x 2 4 x

2 2 2A 2 2 (x 2)(4 x) A 2 (x 2) (4 x) A 4 (BT Csi)

VT 2 vi 2 x 4

Du bng xy ra khi v chkhi x2 = 4x x = 3

Mt khc VP = 2 2x 6x 11 (x 3) 2 2 , du bng xy ra khi v chkhi x = 3

Suy ra phng trnh cho tng ng vi h2

x 2 4 x 2x 3

x 6x 11 2

Vy x = 3 l nghim duy nht ca phng trnh

V d4:Gii phng trnh

3x7x3 2 + 4x3x2 = 2x 2 + 1x5x3 2 (1)

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Gii.

Vit 3x7x3 2 = )2x(21x5x3 2

4x3x2 = )2x(32x 2

Vy (1)

01x5x3

02x02x

2

2 x = 2.

V d5:Gii phng trnh

2 2 23x 6x 7 5x 10x 14 4 2x x (1)

Gii.

2 2 2(1) 3(x 1) 4 5(x 1) 9 5 (x 1)

VT(1) 5, VP(1) 5, x

VT(1) 5(1) x 1 0 x 1

VP(1) 5

Vy x = -1 l nghim duy nht ca phng trnh

V. GII BT PHNG TRNH BNG CCH NHN LNG LIN HP

MT SV D:

V d1:Gii bt phng trnh

21 1 4x3

x

Bng cch nhn lng lin hp bt phng trnh tng ng

2

2

2 2

4x3

x 1 1 4x

x 0 x 0

4x 3 3 1 4x 3 1 4x 4x 3

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21 4x c ngha th1 1

x2 2

. V x 1

24x3< 0

Do (1),(2)x 0

1 1x

2 2

. Tp nghim 1 1

S ; \ 02 2

V d2:Gii bt phng trnh

2

12x 82x 4 2 2 x

9x 16

(1)

Bng cch nhn lng lin hp bt phng trnh tng ng

22

6x 4 2(6x 4)(3x 2) 9x 16 2 2x 4 2 2 x 0

2x 4 2 2 x 9x 16

(2)

Li thc hin php nhn lin hp

2 2

2 2

2 2

(2) (3x 2) 9x 16 4 12 2x 4 8 2x 0

(3x 2) 9x 8x 32 16 8 2x 0

(3x 2) x 2 8 2x x 2 8 2x 0 (3)

28 2x c ngha th -2 x 2. Do 2x 2 8 x 2 8 2x 0 nn

2

2 2

(3) (3x 2) x 2 8 2x 0

3x 2 0 3x 2 0 (I) (II)

x 2 8 2x 0 x 2 8 2x 0

Gii (I)4 2

x 23

Gii (II) 2 2 2

2x 0 0 x< 2 2

2 x 0 0 x< 2 x

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V. Tslng gic ca cung lng gic

A- L thuyt.1. Cng thc lng gic c bn.

sin k2 sin

cos( k2 ) costan k tan

cot k cot

2 2sin cos 1

sintan , kcos 2

coscot , k

sin

2

2

2

2

11 tan

cos

1 1 cotsin

1cot , k

tan 2

2.Gi trlng gic ca cc gc (cung) c lin quan c bit.

sin(- ) = sin

cos(- )= - cos

tan tancot cot

cos2

sin

sin2

cos

tan cot2

cot tan2

sin(- ) = - sin

cos(- ) = cos

tan(-

) = - tan

cot(- ) = - cot

sin(+ ) = - sin

cos(+ )= - cos

tan tancot cot

3. Mt scng thc lng gic thng gp.

a) Cng thc cng.

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cos( ) cos .cos -sin .sin

cos( ) cos .cos sin .sin

sin( ) sin .cos cos .sin

sin( ) sin .cos cos .sin

tan tantan( )

1 tan tan

tan tantan( )

1 tan tan

b) Cng thc nhn i.

2 2 2 2

2

sin2 2sin .cos

cos2 cos sin 2cos 1 1 2sin

2tantan2

1 tan

c) Cng thc hbc.

2 21 cos2 1 cos2sin ; cos2 2

d) Cng thc bin i tch thnh tng.

1cos .cos cos cos

2

1sin .sin cos cos

2

1sin .cos sin sin

2

e) Cng thc bin i tng thnh tch.

cos cos 2cos cos2 2

cos cos 2sin sin2 2

sin sin 2sin cos

2 2sin sin 2cos cos

2 2

4. Bng gi trlng gic ca cc cung c bit.

Gc

00 300 450 600 900 1200 1350 1500 1800

06

4

3

2

2

3

3

4

5

6

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sin 01

2 2

2

3

2 1

3

2

2

2

1

2 0

cos 1 3

2

2

2

1

2 0

1

2

2

2

3

2 -1

tan 01

3

1 3 kx - 3 -1 1

3

0

cot kx 3 11

3 0

1

3 -1 3 kx

* Ch :

cos 0 khi im ngn ca cung thuc gc phn tthI v IVcos 0 khi im ngn ca cung thuc gc phn tthII v IIIsin 0 khi im ngn ca cung thuc gc phn tthI v IIsin 0 khi im ngn ca cung thuc gc phn tthIII v IV

B- Cc bi tp p dng.

Bi 1:Tnh cc gi trlng gic ca gc bit.a) sin=

2 3v

Ôn Tập Toán 10 Phần Đại Số

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