Chương 2 Giải phương trình Đại Số và phương trình Siêu Việt pot
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Bài 1:
= 10
-3
Câu1:
125.1sin xx
1,5.1 x
125.1sin)( xxxf
1,5.1 x
91
2ln
10
5.11
ln
1
2ln
ln
3
ab
nh
025.1
2
15.1
2
)(
0371242.0)5.1()(
0283529.0)1()(
ba
c
faf
fbf
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
-1.5
-1
-1.25
+
+
-
2
-1.25
-1
-1.125
-
+
-
3
-1.25
-1.125
-1.1875
-
+
-
4
-1.25
-1.1875
-1.217875
+
+
-
5
-1.21875
-1.1875
-1.203125
-
+
-
6
-1.21875
-1.203125
-1.210937
+
+
-
7
-1.2109375
-1.203125
-1.207031
+
+
-
8
-1.207031
-1.203125
-1.205078
+
+
-
9
-1.207031
-1.205078
-1.206054
+
+
-
206054.1x
:
: C
*
X
*
-6
Câu 2:
02cos xx
1,0x
xxxf 2cos)(
101
2ln
10
01
ln
3
h
5.0
2
1
2
)(
10.09172,6)1.2cos(1)(
1)0.2cos()(
4
ba
cf
bf
af
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
0
1
0.5
+
-
+
2
0
0.5
0.25
-
-
+
3
0.25
0.5
0.375
-
-
+
4
0.375
0.5
0.4375
+
-
+
5
0.375
0.4375
0.40625
-
-
+
6
0.40625
0.4375
0.421875
-
-
+
7
0.421875
0.4375
0.429688
+
-
+
8
0.421875
0.429688
0.425781
-
-
+
9
0.425781
0.429688
0.427735
+
-
+
10
0.425781
0.427735
0.426758
-
-
+
*
C
*
6
101
*
10.907226.1
2
207031.1205078.1
2
n
ab
XC
7
111
*
10.541015.9
2
425781.0427735.0
2
n
ab
XC
Câu 3:
0)25.0( xtgx
2,5.1x
Xét
)25.0()( xtgxxf
2,5.1x
91
2ln
10
5.12
ln
1
2ln
ln
3
ab
h
:
0351420.12)25.075.1(75.1)(
75.1
2
0520380.7)25.02(2)(
0509570.1)25.05.1(5.1)(
tgcf
ba
c
tgbf
tgaf
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
1.5
2
1.75
-
-
+
2
1.75
2
1.875
+
-
+
3
1.75
1.875
1.8125
-
-
+
4
1.8125
1.875
1.843750
+
-
+
5
1.8125
1.843750
1.828125
+
-
+
6
1.8125
1.828125
1.820313
-
-
+
7
1.820313
1.828125
1.820781
-
-
+
8
1.820781
1.828125
1.824453
+
-
+
9
1.820781
1.824450
1.826617
+
-
+
*
6
101
*
10.583007.3
2
820781.1824450.1
2
n
ab
CX
Câu 4:
2
)1( xxtg
1,0x
2
)1()( xxtgxf
1,0x
101
2ln
10
1
ln
1
2ln
ln
3
ab
h
0851420.11)(
5.0
2
0185040.64)11()(
0557408.10)1()(
cf
ba
c
tgbf
tgaf
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
0
1
0.5
+
+
-
2
0.5
1
0.75
-
+
-
3
0.5
0.75
0.625
-
+
-
4
0.5
0.625
0.5625
+
+
-
5
0.5625
0.625
0.593750
-
+
-
6
0.5625
0.593750
0.578125
-
+
-
7
0.5625
0.578125
0.570313
+
+
-
8
0.570313
0.578125
0.574219
-
+
-
9
0.570313
0.574219
0.572266
-
+
-
10
0.570313
0.572266
0.571290
-
+
-
571290.0x
*
6
111
*
10.042481.1
2
570313.0572266.0
2
n
ab
CX
Câu 5:
02
3
xx
4,3x
2)(
3
xxxf
4,3x
101
2ln
10
34
ln
3
h
018294.0)5.3()(
5.3
2
43
2
0412599.0)4()(
0442250.0)3()(
fcf
ba
C
fbf
faf
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
3
4
3.5
-
-
+
2
3.5
4
3.75
+
-
+
3
3.5
3.75
3.625
+
-
+
4
3.5
3.625
3.5625
+
-
+
5
3.5
3.5625
3.531250
-
-
+
6
3.53125
3.5625
3.546875
+
-
+
7
3.53125
3.546875
3.539073
+
-
+
8
3.53125
3.539063
3.535156
+
-
+
9
3.53125
3.535146
3.533203
+
-
+
10
3.53125
3.533203
3.532227
+
-
+
532227.3x
*
6
11
*
10.1
2
353125533203.3
CX
Câu 6:
0sin4
2
xx
3,1x
xxxf sin4)(
2
3,1x
111
2ln
10
2
ln
3
h
362810.02sin42)(
2
2
31
2
0435520.83sin43)3()(
0365884.21sin41)1()(
2
2
2
cf
ba
c
fbf
faf
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
1
3
2
+
-
+
2
1
2
1.5
-
-
+
3
1.5
2
1.75
-
-
+
4
1.75
2
1.875
-
-
+
5
1.875
2
1.9375
+
-
+
6
1.875
1.9375
1.90625
-
-
+
7
1.906250
1.9375
1.921875
-
-
+
8
1.921875
1.9375
1.929688
-
-
+
9
1.929688
1.9375
1.933594
-
-
+
10
1.933594
1.9375
1.935547
+
-
+
935547.1x
*
6
111
10.2
2
933594.19375.1
2
*
n
ab
CX
Câu 7: lnx 3xsinx + 2 = 0 ; x [0.1;0.7]
101
2ln
)
10
1.07.0
ln(
3
n
f(a) = f(0.1) = ln0.1 3.0,1sin0.1 + 2 = -0.332553 < 0
f(b) = f(0.7) > 0
0616407.0)(
4.0
2
1.07.0
2
xf
ba
c
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
0.1
0.7
0.4
+
-
+
2
0.1
0.4
0.25
+
-
+
3
0.1
0.25
0.175
+
-
+
4
0.1
0.175
0.1375
-
-
+
5
0.1375
0.175
0.15625
+
-
+
6
0.1375
0.15625
0.14675
+
-
+
7
0.1375
0.14675
0.142188
-
-
+
8
0.142188
0.14675
0.144469
+
-
+
9
0.142188
0.144469
0.143328
-
-
+
10
0.143328
0.144469
0.143899
-
-
+
x = 0.143899
GX
*
trình
6
111
*
10.132.57
2
143328.0144469.0
2
n
ab
CX
Câu 8: f (x) = x
2
- 4sinx - 5 ;
3,2x
T
101
2ln
ln
ab
n
0)3()(
0)2()(
fbf
faf
143889.1)5.2()(
5.2
2
32
2
fxf
ba
c
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
2
3
2.5
-
-
+
2
2.5
3
2.75
+
-
+
3
2.5
2.75
2.625
-
-
+
4
2.625
2.75
2.6875
+
-
+
5
2.625
2.6875
2.65625
+
-
+
6
2.625
2.65625
2.640625
+
-
+
7
2.625
2.640625
2.632813
-
-
+
8
2.632813
2.640625
2.636719
+
-
+
9
2.632813
2.636719
2.634766
-
-
+
10
2.634766
2.636719
2.635742
+
-
+
x = 2.635742
G
*
6
1111
*
10.536.9
2
634766.2636719.2
2
ab
CX
Cau 9:
13sin)1ln()( xxxxxf
2,1.1x
Tí
101
2ln
)ln(
ab
h
0)2()(
0)1.1()(
fbf
faf
55.1
2
21.1
2
ba
c
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
1.1
2
1.55
+
-
+
2
1.1
1.55
1.325
+
-
+
3
1.1
1.325
1.2125
-
-
+
4
1.2125
1.325
1.26875
+
-
+
5
1.26875
1.325
1.310375
+
-
+
6
1.26875
1.310375
1.289563
+
-
+
7
1.26875
1.289563
1.279156
+
-
+
8
1.26875
1.279156
1.273953
+
-
+
9
1.26875
1.273953
1.271352
+
-
+
10
1.26875
1.271352
1.270051
+
-
+
G
*
6
111
*
10.27.1
2
26875.1271352.1
2
n
ab
XC
Câu 10:
02cos3)1ln(
x
exxx
2,1x
2cos3)1ln()(
x
exxxxf
101
2ln
10
12
ln
3
h
492913.2)2()(
428349.1)1()(
fbf
faf
5.1
2
ba
c
403401.05.1)( fcf
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
1
2
1.5
+
+
-
2
1.5
2
1.75
-
+
-
3
1.5
1.75
1.625
-
+
-
4
1.5
1.625
1.5625
-
+
-
5
1.5
1.5625
1.53125
-
+
-
6
1.5
1.53125
1.515625
+
+
-
7
1.515625
1.53125
1.523438
-
+
-
8
1.515625
1.523438
1.519531
+
+
-
9
1.519531
1.523438
1.521485
+
+
-
10
1.521482
1.523438
1.522461
-
+
-
*
7
11
10.54.9
2
001953.10009765.1
*
CX
Câu 11:
02015
3
12
8
xx
x
x
1,5.1x
2015
3
12
)(
8
xx
x
x
xf
1,5.1x
91
2ln
)
10
5.11
ln(
3
h
05.4)1()(
0462240.27)5.1()(
fbf
faf
0893968.3)25.1()(
25.1
2
15.1
2
fcf
ba
c
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
-1.5
-1
-1.25
-
-
+
2
-1.25
-1
-1.125
+
-
+
3
-1.25
-1.125
-1.1875
-
-
+
4
-1.1875
-1.125
-1.15625
+
-
+
5
-1.1875
-1.15625
-1.175875
-
-
+
6
1.171875
-1.15625
-1.164062
-
-
+
7
-1.164062
-1.15625
-1.60156
+
-
+
8
-1.164062
-1.160156
-1.162109
+
-
+
9
-1.164062
-1.162109
-1.163086
-
-
+
-
*
6
101
*
10.9.1
2
164062.1163086.1
2
n
ab
CX
Câu 12:
02
log1
sin1
125.1
2
x
x
x
6.0,55.0x
2
log1
sin1
)(
125.1
2
x
x
x
xf
61
2ln
10
55.06.0
ln
3
h
0575.0
2
6.055.0
2
0123945.0)6.0()(
0817336.0)55.0()(
ba
c
fbf
faf
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
0.55
0.6
0.575
+
+
-
2
0.575
0.6
0.5875
+
+
-
3
0.5875
0.6
0.593750
-
+
-
4
0.5875
0.593750
0.590625
-
+
-
5
0.5875
0.590625
0.589063
+
+
-
6
0.589063
0.590625
0.589844
+
+
-
*
5
71
*
10.22.1
2
551562.0550781.0
2
n
ab
XC
Câu 13:
04
)1.1sin(
32
7
)12ln(
x
x
x
5.1;1x
4
)1.1sin(
32
)(
7
)12ln(
x
x
xf
x
5.1;1x
91
2ln
10
115.1
ln
3
h
212446.0)25.1()(
25.1
2
5.11
076037.2)5.1()(
443867.1)1()(
fcf
c
fbf
faf
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
1
1.5
1.25
+
+
-
2
1.25
1.5
1.375
-
+
-
3
1.25
1.375
1.3125
-
+
-
4
1.25
1.3125
1.281250
+
+
-
5
1.281250
1.3125
1.296875
-
+
-
6
1.281250
1.296875
1.289063
-
+
-
7
1.281250
1.289063
1.285156
-
+
-
8
1.281250
1.285156
1.283203
-
+
-
9
1,283203
1.285156
1.284180
-
+
-
*
6
101
*
10.2
2
283203.1285156.1
2
n
ab
CX
Câu 14:
09.11
3
)1ln(
5
2
2
2
x
xx
x
15.1;1x
9.11
3
)1ln(
)(
5
2
2
2
x
xx
x
xf
15.1;1x
81
2ln
10
115.1
ln
3
h
009455.0)075.1()(075.1
2
115.1
0036420.0)15.1()(
0017649.0)1()(
fcfc
fbf
faf
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
1
1.15
1.075
+
-
+
2
1
1.075
1.0375
-
-
+
3
1.0375
1.075
1.05625
+
-
+
4
1.0375
1.05625
1.046875
-
-
+
5
1.046875
1.05625
1.051552
+
-
+
6
1.046875
1.051552
1.049214
+
-
+
7
1.046875
1.049214
1.048044
-
-
+
8
1.048044
1.049214
1.048629
-
-
+
*
:
6
9
*
10.1098251.2
2
048044.1049214.1
CX
Câu 15:
070
)1cos(2
53
10
2
2
x
x
xx
75.1;5.1x
70
)1cos(2
53
)(
10
2
2
x
x
xx
xf
75.1;5.1x
91
2ln
10
5.175.1
ln
3
h
925718.55)(625.1
2
75.15.1
2
373119.197)75.1()(
068912.15)5.1()(
cf
ba
c
fbf
faf
n
a
n
b
n
c
n
f(c
n
)
f(a
n
)
f(b
n
)
1
1.5
1.75
1.625
-
+
-
2
1.5
1,625
1.5625
-
+
-
3
1.5
1.5625
1.53125
+
+
-
4
1.53125
1.5625
1.546875
-
+
-
5
1.53125
1.546875
1.539063
-
+
-
6
1.53125
1.539063
1.535156
-
+
-
7
1.53125
1.535156
1.533203
+
+
-
8
1.533203
1.535156
1.534180
+
+
-
9
1.534180
1.535156
1.534668
+
+
*
6
10
*
10.1
2
534180.1535156.1
CX
Bài 2: Gi
v chính xác là = 10
-5
.
Câu 1: x
3
=0
=
i t:
max
= 0.029987 = q < 1 thi t.
Chn:
o
=
n+1
=(x
n
) =
x
1
x
2
x
3
x
4
x
5
x
6
x
7
Gia tri
x
n+1
1.357209
1.333861
1.325884
1.324939
1.324760
1.324726
1.324719
| x
7
- x
6
| = 7.10
-6
< 10
-5
là nghim g
*
là nghim g
66
67
*
7
10.210.7
29987.01
29987.0
1
xx
q
q
xx
Câu 2:
=
n+1
=(x
n
) =
n+1
1.946423
1.946846
1.947013
1.947080
1.947106
1.947116
=10
-6
< 10
-5
*
12
66
1112
*
12
10.649071.010.
393598.01
393598.0
1
xx
q
q
xx
Câu 3:
4
'
3
34
34
12
)(
)(
4
2
42
042
x
x
x
x
x
xx
xx
n+1
1.767059
1.875299
1.918610
1.935827
1.942651
1.945353
Nên
là nghim g
: Gi s
là nghim chính xác c
Gi s
là nghim g
Câu 4:
0tgxx
1;2.0x
xtgx
2
xtgx
2
)(
suy ra
xtgx
2'
1)(
0)(1).(2)(
1)(
2'
2
xtgxtgxh
xtgxh
Suy ra hàm không hi t trong khong
1;2.0x
1;2.0x
0425519.3)()(
'
max
xxh
Câu 5:
,2
f(x)=
=>
=
=h(x)
= h(0)=0.25<1
=>
=
=f(x)=
= 3.641593
= 3.626049
= 3.626996
= 3.626939
= 3.626942
= 3
Câu 6:
x=
f(x)=
.
.
.
.
.
.
6
65
10
xx
.
cau11:
x=
f(x)=
.
.
.
.
.
.
.
trình
Câu 7:
2
1,75 3
x
x
; [-1; -0,5]
0375.1)(
2
x
xxf
2
2
log( 1,75)
3 1,75
log3
x
x
xx
Xét
2
5
log( 1,75)
, [ 1, 0,5]; 10
log3
x
x
Ta có :
22
22
2ln3( 1,75) 4 ln3
0; [ 1; 0,5]
[( 1,75)ln3]
xx
x
x
-1;-0,5]
ax
( 1) 0,180543
m
hh
|
|
max
=0,180543= q<1
0
0,75
2
ab
x
Mà
)
2
log( 1,75)
log3
n
x
1
x
2
0
log( 1,75)
0,763080
log3
x
=
) =-0,770837
3
5
6
7
8
9
10
11
12
13
14
15
16
0,775469
0,779915
0,780919
0,781524
0,781888
0,782108
0,782240
0,782320
0,782368
0,782397
0,782414
0,782425
0,782431
x
x
x
x
x
x
x
x
x
x
x
x
x
65
16 15
6 10 10xx
* 5 6
16 16 15
0,180542
6 10 1,321922 10
1 1 0,180543
q
x x x x
q
Câu 8:
Ta có:
>0 ;
lu[2;3]
=
=
Mà
1n
x
=
)
=
)
0
4 ln2x
65
13 12
6 10 10xx
* 5 5
13 13 12
0,5
6 10 6 10
1 1 0,5
q
x x x x
q
2
3
4
5
6
7
8
9
10
11
12
13
2,390562
2,435324
2,416773
2,424420
2,421261
2,422564
2,422062
2,422248
2,422157
2,422195
2,422179
2,422185
x
x
x
x
x
x
x
x
x
x
x
x
Câu 9:
3
3
3
2
3
1 0; [1;2]
1
( ) 1 0 1 ( ) '( )
6 ( 1)
x x x
f x x x x x x x
xx
ax
'( ) 0,104993 1 '( ) 1; [1;2]
m
x q x x
Mà
1n
x
=
)
3
1
n
x
=
)
3
0
1 1,305449x
2
3
4
5
6
65
65
1,289173
1,287738
1,287611
1,287600
1,28799
10 10
x
x
x
x
x
xx
6
x
G
*
x
6
x
66
6 6 5
0,104993
* 10 0,11731 10
1 1 0,104993
q
x x x x
q
Câu 10:
03
4sin
xe
x
2;1x
3)(
4sin
xexf
x
4
1
3
s
ex
4
1
3)(
s
ex
4
3
sinsin'
3 cos.
4
1
)(
xx
eexx
0
)3.(4
cos.
4
3
)3).(sin(cos
)3.()3.(4
.cos.
4
3
)3.()3)(sin(cos
)3.(4
)3(.cos.
4
3
)3)(sin(cos
3.4
)3(.cos.
4
3
3.cos.sin.
)3(4
.cos.1
)(
3 cos.
4
1
)(
4
7
sin
sin2sin
4
1
sin
2
3
sin
sin
4
1
sin
4
3
sin2sin
2
3
sin
4
1
sinsin
4
3
sin2sin
2
4
3
sin
4
1
sinsin
4
3
sinsin2sin
4
3
sin
sin
4
3
sinsin
x
xx
xx
xxxx
x
xxxx
x
xxxxx
x
x
xx
e
xexxe
ee
exeexxe
e
eexexxe
e
eexeexex
e
ex
xh
eexxh
1089455.0(max)
0089455.0)1((max)
hx
hh
Tính x
1
, x
2
n
546326.1;546324.1
545921.1)3(
5.1
2
21
)3(
32
4
1
sin
1
0
4
1
sin
1
0
xx
ex
x
ex
x
x
n
n
3
