Ôn Tập Toán 10 Phần Đại Số
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Transcript of Ôn Tập Toán 10 Phần Đại Số
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8/12/2019 n Tp Ton 10 Phn i S
1/21
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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8/12/2019 n Tp Ton 10 Phn i S
2/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
3/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
4/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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)
Bi 12:Gii cc phng trnh sau:
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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8/12/2019 n Tp Ton 10 Phn i S
5/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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
Bi 4:Gii cc phng trnh sau:
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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8/12/2019 n Tp Ton 10 Phn i S
6/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
7/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
8/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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
Bi 8:Gii cc bt phng trnh sau:
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)
Bi 9:Gii cc bt phng trnh sau:
a) 3x 5 2 e)2 x
2x 1
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8/12/2019 n Tp Ton 10 Phn i S
9/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
10/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
11/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
12/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
13/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
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n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
15/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
16/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
17/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
18/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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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8/12/2019 n Tp Ton 10 Phn i S
19/21
n tp ton 10 phn i s (lp 10C3)
GV son: PhmTun Khng Trang
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