Chương 2
1/ xsinx=1; x

F(a)=f(1)=-0.158529<0
F(b)=f(2)=0.81859>0
F(c1)=f[(a+b)/2]=f(1.5)=0.4>0
F(c2)=ff(1.25)=0.18>0
F(c3)= f(1.125)=0.015>0
F(c4)= f(1.0625)=-0.07<0
F(c5)= f(1.09375)=-0.02<0
F(c6)= f(1.109375)=-6.6x10-3<0
F(c7)= f(1.117188)=4.2x10-3>0
F(c8)= f(1.113247)=-1.2<0
F(c9)= f(1.1152175)=1.4x10-3>0
F(c10)= f(1.114233)
Vậy nghiệm của PT là: 1.114233


2/

F(a)=f(0)=-1<0
F(b)=f(1)=0.4596>0
F(c1)=f[(a+b)/2]=f(0.5)=-0.1<0
F(c2)=f(0.75)=-0.1<0
F(c3)= f(0.625)=-0.02<0
F(c4)= f(0.6875)=0.05>0
F(c5)= f(0.65625)=0.01>0
F(c6)= f(0.640625)=-1.3x10-3<0
F(c7)= f(0.645438)=8.2x10-3>0
F(c8)= f(0.644532)=3.4x10-3>0
F(c9)= f(0.642579)=1.05x10-3>0

F(c10)= f(0.641602)
Vậy nghiệm của PT là: 0.641602
3/ x=tgx;

F(a)=f(4)=2.8>0


F(b)=f(4.5)=-0.1<0
F(c1)=f[(a+b)/2]=f(4.25)=-2.2>0
F(c2)=f(4.375)=1.5>0
F(c3)= f(4.4375)=0.89>0
F(c4)= f(4.46875)=0.44>0
F(c5)= f(4.484375)=0.17>0
F(c6)= f(4.492188)=0.0245>0
F(c7)= f(4.496094)=-0.05<0
F(c8)= f(4.494141)=-0.01482<0
F(c9)= f(4.493165)
Vậy nghiệm của PT là: 4.493165
4/ tg(x+1)=x2  tg(x+1)-x2=0; x ∈ (0,1)

F(a)=f(0)=1.5>0
F(b)=f(1)=-3.1<0
F(c1)=f[(a+b)/2]=f(0.5)=13>0
F(c2)=f(0.75)=-6<0
F(c3)= f(0.625)=-18<0
F(c4)= f(0.5625)=120>0


F(c5)= f(0.59375)=-43<0
F(c6)= f(0.578125)=-136<0

F(c7)= f(0.5703125)=2056>0
F(c8)= f(0.57421875)=-292<0
F(c9)= f(0.572265625)=-680<0
F(c10)= f(0.573242187)
Vậy nghiệm của PT là: 0.573242187
5/

F(a)=f(3)=-0.4<0
F(b)=f(4)=0.412>0
F(c1)=f[(a+b)/2]=f(3.5)=-0.01<0
F(c2)=f(3.75)=0.1>0
F(c3)= f(3.625)=0.08>0
F(c4)= f(3.5625)=0.03>0
F(c5)= f(3.53125)=0.008>0
F(c6)= f(3.515625)=-0.004<0
F(c7)= f(3.523438)=0.001>0
F(c8)= f(3.519532)=-0.001<0


F(c9)= f(3.521485)=0.00009>0
F(c10)= f(3.520509)
Vậy nghiệm của PT là: 3.520509
6/x2=4sinx;

F(a)=f(1)=-2.365<0
F(b)=f(3)=8.435>0
F(c1)=f[(a+b)/2]=f(2)=0.3>0
F(c2)=f(1.5)=-1.7<0
F(c3)= f(1.75)=-0.8<0
F(c4)= f(1.875)=-0.3<0

F(c5)= f(1.9375)=0.01>0
F(c6)= f(1.90625)=-0.1<0
F(c7)= f(1.921875)=-0.06<0
F(c8)= f(1.929688)=-0.02<0
F(c9)= f(1.933594)=-0.0008<0
F(c10)= f(1.935547)=0.009>0
F(c11)= f(1.9345705)
Vậy nghiệm của PT là: 1.9345705


7/ x – 1.3 – cos3x = 0

F(a)=f(0)=-2.3<0
F(b)=f(1)=0.6>0
F(c1)=f[(a+b)/2]=f(0.5)=0.4>0
F(c2)=f(0.25)=-1.7<0
F(c3)= f(0.375)=-1.3>0
F(c4)= f(0.6875)=0.5>0
F(c5)= f(0.71875)=0.2>0
F(c6)= f(0.734375)=0.1>0
F(c7)= f(0.742188)=0.004>0
F(c8)= f(0.746094)=0.007>0
F(c9)= f(0.748047)=-0.001<0
F(c10)= f(0.747071)
Vậy nghiệm của PT là: 0.747071
8/ x2 – 4sinx – 5 = 0; x ϵ [2;3]
N= 10
F(a)=f(2)=-4.6<0
F(b)=f(3)=3.4>0



F(c1)=f[(a+b)/2]=f(2.5)=-1.4<0
F(c2)=f(2.75)=1.03>0
F(c3)= f(2.625)=-0.8<0
F(c4)= f(2.6875)=0.4>0
F(c5)= f(2.65625)=0.18>0
F(c6)= f(2.640625)=0.05>0
F(c7)= f(2.632873)=-0.01<0
F(c8)= f(2.636719)=0.01>0
F(c9)= f(2.634766)=0.0003>0
F(c10)= f(2.6337895)
Vậy nghiệm của PT là: 2.6337895
9/ lnx – 3xsinx + 2 = 0; x ϵ [1,0.25]
N = 10
F(a)=f(1)=-0.52<0
F(b)=f(0.25)=0.42>0
F(c1)=f[(a+b)/2]=f(0.625)=0.4>0
F(c2)=f(0.8125)=0.2>0
F(c3)= f(0.90625)=-0.2<0
F(c4)= f(0.859375)=-0.1<0
F(c5)= f(0.835938)=-0.03<0


F(c6)= f(0.824219)=-0.008<0
F(c7)= f(0.827345)=-0.016<0
F(c8)= f(0.819923)=0.003>0
F(c9)= f(0.823634)=-0.006<0
F(c10)= f(0.821779)
Vậy nghiệm của PT là: 0.821779
10/ lnx – 3xsinx + 2 = 0; x ϵ [0.1,0.7]

N=10
F(a)=f(0.1)=-0.33<0
F(b)=f(0.7)=0.29>0
F(c1)=f[(a+b)/2]=f(0.4)=0.6>0
F(c2)=f(0.25)=0.42>0
F(c3)= f(0.175)=0.16>0
F(c4)= f(0.1375)=-0.04<0
F(c5)= f(0.15625)=0.07>0
F(c6)= f(0.146875)=0.17>0
F(c7)= f(0.142188)=-0.011<0
F(c8)= f(0.144532)=0.003>0
F(c9)= f(0.14336)=-0.003<0
F(c10)= f(0.143946)


Vậy nghiệm của PT là: 0.143946
11/ xln(x-1) – xsin3x +1 = 0. x ϵ [1,2] n=10
Với x = 1 thì f(x) = xln(x-1) – xsin3x +1 vô nghĩa
12/ xln(x+1) – 3xcosex – 2 = 0. x ϵ [1,2] n=10
f(a)

1

1.4

>0

f(b)

2


-2.4

<0

f(c1)

1.5

0.4

>0

f(c2)

1.75

-4.7

<0

f(c3)

1.625

-2.1

<0

f(c4)


1.5625

-0.8

<0

f(c5)

1.53125

-0.17

<0

f(c6)

1.515625

0.125

>0

f(7)

1.5234375

-0.02

<0


f(8)

1.51953125

0.05

>0

f(9)

1.521484375

0.014

>0

f(10)

1.522460938

Vậy nghiệm của PT là: 1.522460938
13/ x3 – 7x2 + 14x – 6 = 0. x ϵ (0,1) v x ϵ (1,3.2) v x ϵ (3.2,4). E=10 -2
Với x ϵ (0,1)  n=7
F(a)=f(1)=-6<0
F(b)=f(2)=0.2>0
F(c1)=f[(a+b)/2]=f(1.5)=-0.625<0
F(c2)=f(0.75)=0.9>0



F(c3)= f(0.625)=0.2>0
F(c4)= f(0.5625)=-0.1<0
F(c5)= f(0.59375)=0.05>0
F(c6)= f(0.578035)=-0.05<0
F(c7)= f(0.585803)
Vậy nghiệm của PT là: 0.585803
Với x ϵ (1,3.2)  n=9
F(a)=f(1)=-0.52<0
F(b)=f(3.2)=0.42>0
F(c1)=f[(a+b)/2]=f(2.1)=0.43>0
F(c2)=f(2.65)=0.22>0
F(c3)= f(2.925)=-0.23<0
F(c4)= f(3.0625)=-0.1<0
F(c5)= f(2.99375)=-0.03<0
F(c6)= f(3.028125)=-0.003<0
F(c7)= f(3.010938)=-0.016<0
F(c8)= f(3.002344)=0.003>0
F(c9)= f(2.998047)=-0.006<0
Vậy nghiệm của PT là: 2.998047
Với x ϵ (3.2,4)  n=7


N=7
F(a)=f(3.2)=-0.112<0
F(b)=f(4)=2>0
F(c1)=f[(a+b)/2]=f(3.6)=0.336>0
F(c2)=f(3.4)=-0.016<0
F(c3)= f(3.5)=0.125>0
F(c4)= f(3.45)=0.046>0
F(c5)= f(3.425)=0.013>0

F(c6)= f(3.4125)=-1.9<0
F(c7)= f(3.4187)
Vậy nghiệm của PT là: 3.4187
14/ x4 – 2x3 – 4x2 +4x + 4 = 0 trong (-2,-1) (0,2) (2,3) (-1,0) e=10-2
Với x ϵ (-2,-1)  n=7
F(a)=f(-2)=12>0
F(b)=f(-1)=-1<0
F(c1)=f[(a+b)/2]=f(-1.5)=0.8125>0
F(c2)=f(-1.25)=-0.9<0
F(c3)= f(-1.375)=-0.2<0
F(c4)= f(-1.4375)=0.1>0
F(c5)= f(-1.40625)=-0.06<0


F(c6)= f(-1.424875)=0.06>0
F(c7)= f(-1.414063)
Vậy nghiệm của PT là: -1.414063
Với x ϵ (0,2)  n=8
F(a)=f(0)=4>0
F(b)=f(2)=-4<0
F(c1)=f[(a+b)/2]=f(1)=3>0
F(c2)=f(1.5)=-0.6<0
F(c3)= f(1.25)=1.2>0
F(c4)= f(1.375)=0.3>0
F(c5)= f(1.4375)=-0.18<0
F(c6)= f(1.40625)=0.06>0
F(c7)= f(1.421875)=-0.06<0
F(c8)= f(1.414063)
Vậy nghiệm của PT là: -1.414063
Với x ϵ (2,3)  n=7

F(a)=f(2)=-4<0
F(b)=f(3)=7>0
F(c1)=f[(a+b)/2]=f(1.5)-0.6875<0
F(c2)=f(2.25)=-4.4<0


F(c3)= f(2.625)=-1.7<0
F(c4)= f(2.8125)=1.6>0
F(c5)= f(2.71875)=-0.2<0
F(c6)= f(2.765625)=0.66>0
F(c7)= f(2.752188)
Vậy nghiệm của PT là: 2.752188
Với x ϵ (-1,0)  n=7
F(a)=f(-1)=-1<0
F(b)=f(0)=4>0
F(c1)=f[(a+b)/2]=f(-0.5)=1.3>0
F(c2)=f(-0.75)=-0.08<0
F(c3)= f(-0.625)=0.5>0
F(c4)= f(-0.6875)=0.23>0
F(c5)= f(-0.71875)=0.06>0
F(c6)= f(-0.734375)=-0.01<0
F(c7)= f(-0.726563)
Vậy nghiệm của PT là: -0.726563
15/ 2 + cos(ex – 2) – ex = 0 x ϵ (0.5,1.5) e=10-3
N=10
F(a)=f(0.5)=1.2>0


F(b)=f(1.5)=-3.2<0
F(c1)=f[(a+b)/2]=f(1)=0.03>0

F(c2)=f(1.25)=-1.4<0
F(c3)= f(1.125)=-0.6<0
F(c4)= f(1.0625)=-0.2<0
F(c5)= f(1.03125)=-0.1<0
F(c6)= f(1.015625)=-0.03<0
F(c7)= f(1.007813)=-0.0008<0
F(c8)= f(1.003907)=0.01>0
F(c9)= f(1.00586)=0.008>0
F(c10)=f(1.006837)
Vậy nghiệm của PT là: 1.006837
16/ x – 2-x . x ϵ (0,1) e=10-3
N=10
F(a)=f(0)=-1<0
F(b)=f(1)=0.5>0
F(c1)=f[(a+b)/2]=f(0.5)=-0.2<0
F(c2)=f(0.75)=0.15>0
F(c3)= f(0.625)=-0.02<0
F(c4)= f(0.6875)=0.6>0


F(c5)= f(0.65625)=0.02>0
F(c6)= f(0.640625)=-0.0008<0
F(c7)= f(0.648438)=0.01>0
F(c8)= f(0.644532)=0.004>0
F(c9)= f(0.642579)=0.002>0
F(c10)=f(0.641602)
Vậy nghiệm của PT là: 0.641602
17/ ex – x2 + 3x - 2 . x ϵ (0,1) e=10-3
N= 10
F(a)=f(0)=-1<0

F(b)=f(1)=2.7>0
F(c1)=f[(a+b)/2]=f(0.5)=0.89>0
F(c2)=f(0.25)=-0.02<0
F(c3)= f(0.375)=0.43>0
F(c4)= f(0.3125)=0.2>0
F(c5)= f(0.28125)=0.08>0
F(c6)= f(0.265625)=0.03>0
F(c7)= f(0.257813)=0.001>0
F(c8)= f(0.253907)=-0.01<0
F(c9)= f(0.25586)=-0.006<0


F(c10)=f(0.2568337)
Vậy nghiệm của PT là: 0.2568337
18/ 2xcos2x – (x + 1)2 = 0 trong (-3,-2) (-1,0) e=10-3
Với x ϵ (-3,-2)  n=10
F(a)=f(-3)=-9.7<0
F(b)=f(-2)=1.6>0
F(c1)=f[(a+b)/2]=f(-2.5)=-3.6<0
F(c2)=f(-2.25)=-0.6<0
F(c3)= f(-2.125)=0.6>0
F(c4)= f(-2.1875)=0.03>0
F(c5)= f(-2.21875)=-0.28<0
F(c6)= f(-2.203125)=-0.11<0
F(c7)= f(-2.195313)=-0.04<0
F(c8)= f(-2.125)=-0.0009<0
F(c9)= f(-2.1875)=0.01>0
F(c10)=f(-2.190581)
Vậy nghiệm của PT là: -2.190581
Với x ϵ (-1,0)  n=10

F(a)=f(-1)=0.8>0
F(b)=f(0)=-1<0


F(c1)=f[(a+b)/2]=f(-0.5)=-0.79<0
F(c2)=f(-0.75)=-0.16<0
F(c3)= f(-0.875)=1.2>0
F(c4)= f(-0.8125)=0.05>0
F(c5)= f(-0.78125)=-0.06<0
F(c6)= f(-0.796875)=-0.004<0
F(c7)= f(-0.804688)=0.02>0
F(c8)= f(-0.800782)=0.009>0
F(c9)= f(-0.798829)=0.002>0
F(c10)=f(-2.797852)
Vậy nghiệm của PT là: -2.797852
19/ xcosx – 2x2 + 3x – 1 = 0 trong (0.2,0.3) (1.2,1.3) e=10-3
Với x ϵ (0.2,0.3)  n = 7
f(a)

0.2

-0.2

<0

f(b)

0.3

0.006


>0

f(c1)

0.25

-0.13

<0

f(c2)

0.275

-0.06

<0

f(c3)

0.2875

-0.02

<0

f(c4)

0.290375


-0.01

<0

f(c5)

0.296875

-0.001

<0

f(c6)

0.298438

0.002

>0


f(7)

0.297657

Vậy nghiệm của PT là: 0.297657
Với x ϵ (1.2,1.3)  n = 7
f(a)


1.2

0.15

>0

f(b)

1.3

-0.13

<0

f(c1)

1.25

0.01

>0

f(c2)

1.275

-0.05

<0


f(c3)

1.2625

-0.01

<0

f(c4)

1.25625

0.001

>0

f(c5)

1.259375

-0.008

<0

f(c6)

1.257813

-0.003


<0

f(7)

1.2570315

Vậy nghiệm của PT là: 1.2570315
20/ x3 + x – 4 = 0 trong (1.4) e=10-2  n = 9
f(a)

1

-2

<0

f(b)

4

64

>0

f(c1)

2.5

14.125


>0

f(c2)

1.75

3.1

>0

f(c3)

1.375

-0.025

<0

f(c4)

1.5625

1.3

>0

f(c5)

1.46875


0.6

>0

f(c6)

1.421875

0.2

>0

f(7)

.1398438

0.1

>0

f(8)

1.386719

0.05

>0


f(9)


1.3808595

Vậy nghiệm của PT là: 1.3808595
21/ x – sinx = 1  x – sinx – 1 = 0 trong (1.2) e=10-3  n = 10
f(a)

1

-0.84

<0

f(b)

2

0.09

>0

f(c1)

1.5

-0.9

<0

f(c2)


1.75

-0.2

<0

f(c3)

1.875

-0.07

<0

f(c4)

1.9375

0.003

>0

f(c5)

1.90625

-0.03

<0


f(c6)

1.921875

-0.01

<0

f(7)

1.929688

-0.006

<0

f(8)

1.933594

-0.001

<0

f(9)

1.935547

0.001


>0

f(10)

1.934571

Vậy nghiệm của PT là: 1.934571
22/ tg(x + 1) = x2  tg(x + 1) – x2 = 0 trong (0.1) e=10-3  n = 10
f(a)

0

1.5

>0

f(b)

1

-3.1

<0

f(c1)

0.5

13.8


>0

f(c2)

0.75

-6.08

<0

f(c3)

0.625

-18.8

<0

f(c4)

0.5625

120.2

>0

f(c5)

0.59375


-43

<0


f(c6)

0.578125

-136.7

<0

f(7)

0.570313

2066.5

>0

f(8)

0.574219

-292.4

<0


f(9)

0.572266

-680.7

<0

f(10)

0.5712895

Vậy nghiệm của PT là: 0.5712895