Bài tập Giải tích số (Có lời giải)

f(x) = 2
x
x
k
T
4
(x)
x
k
= cos
π + k2π
2n
, n = 4, k = 0, 1, 2, 3.
x
1
= cos
Π
8
≈ 0, 924 → f (x
1
) = 1, 897
x
2
= cos
3Π
8
≈ 0, 383 → f (x
2
) = 1, 304
x
3

= cos
5π
8
≈ −0, 383 → f (x
3
) = 0, 767
x
4
= cos
7π
8
≈ −0, 924 → f (x
4
) = 0, 527
P
3
(x) =
3

i=0
f(x
i
)P
i
(x)
= 1, 897
(x −0, 383)(x + 0, 383)(x + 0.924)
(0, 924 − 0, 383)(0, 924 + 0, 383)(0, 924 + 0, 924)
+ 1, 304
(x −0.924)(x + 0, 383)(x + 0, 924)

(0, 383 − 0.924)(0, 383 + 0, 383)(0, 383 + 0, 924)
+ 0.767
(x −0.924)(x −0, 383)(x + 0.924)
(−0, 383 − 0.924)(−0, 383 − 0, 383)(−0, 383 + 0, 924)
+ 0, 527
(x −0.924)(x −0, 383)(x + 0, 383)
(−0, 924 − 0, 924)(−0, 924 − 0, 383)(−0, 924 + 0, 383)
= 0, 86x
3
+ 0, 25x
2
+ 0, 58x − 1, 1
f(x)
ξ
f(x) = cos x [0,
π
2
], ξ =
π
12
f(x) = x cos x [0,
π
2
], ξ =
π
6
{0,
π
4
,

π
2
}
π
4
π
2
√
(2)
2
P
2
(x) =
(x −
π
2
)(x −
π
4
)
π
4
×
π
2
−
√
2
2
×

(x −
π
2
)x
π
4
×
π
4
= −0, 3357x
2
− 0, 1092x + 1
⇒ P
2
(
π
12
) =
8(1 −
√
2)
π
2
12
2
+ 2(2
√
2 −3)
π
2

12
+ π
2
π
2
=
40
√
2 + 80
12
2
= 0, 948
f(x) = cos x P
2
(x) =
−0, 3357x
2
− 0, 1092x + 1
π
12
| P
2
(
π
12
) −f(
π
12
) |≤
M

3
3!
| ω
3
(
Π
12
) |
M
3
= max
[0,
Π
2
]
| y
(3)
(x) |= max
[0,
Π
2
]
| sin x |= 1
ω
3
(
Π
12
) =| (
Π

12
− 0)(
Π
12
−
Π
4
)(
Π
12
−
Π
2
) |
⇒| P
2
(
π
12
) −f(
π
12
) |≤ 0, 0299
cos (
Π
12
) = 0, 9484 ± 0, 0299
0,
π
4

,
π
2
π
4
π
2
√
(2)
8
π
P
2
(x) = −
√
2
8
×
(x −
π
2
)x
π
4
×
π
4
= −
2
√

2
π
x
2
+
√
2x
⇒ P
2
(
π
6
) =
√
2
9
π = 0, 4936
π
6
| P
2
(
π
6
) − f (
π
6
) |≤
M
3

3!
| ω
3
(
Π
6
) |≤ 0, 0718
f(x) = x cos x P
2
(x) =
−
2
√
2
π
x
2
+
√
2x
S
n
= 1 + 2
3
+ 3
3
+ + n
3
S
n

= S
n+1
− S
n
= (n + 1)
3

2
S
n
= S
n+1
− S
n
= (n + 2)
3
− (n + 1)
3
= 3n
2
+ 9n + 7

3
S
n
= 
2
S
n+1
− 

2
S
n
= 6n + 12

4
S
n
= 
3
S
n+1
− 
3
S
n
= 6 = const
S
n
S
n
S
n

2
S
n

3
S

n

4
S
n
x
0
= 1
⇒ S
n
= 1 + 8(n −1) +
19
2
(n − 1)(n −2) +
18
3!
(n − 1)(n −2)(n −3)
+
6
3!
(n − 1)(n −2)(n −3)(n −4)
=
n
4
+ 2n
3
+ n
2
4
=

n
2
(n + 1)
2
4
S
n
=
n
2
(n+1)
2
4
A =
x
2
+6x+1
(x+1)(x−1)(x−4)(x−6)
B =
3x
2
+x+1
(x−1)(x−2)(x−3)
f(x) = x
2
+ 6x + 1 −1, 1, 4, 6
−1 −4
x
0
= −1

f(x) = −4 + 6(x + 1) + (x − 1)(x + 1)
⇒ A =
4
(x+1)(x−1)(x−4)(x−6)
+
6
(x−1)(x−4)(x−6)
+
1
(x−4)(x−6)
A =
4
(x + 1)(x −1)(x −4)(x −6)
+
6
(x − 1)(x −4)(x −6)
+
1
(x − 4)(x −6)
g(x) = x
2
+ 6x + 1 1, 2, 3
f(x) 
2
f(x)
x
0
= 1
g(x) = 5 + 10(x − 1) +
6

2!
(x − 1)(x −2) = 5 + 10(x −1) + 3(x −1)(x −2)
⇒ B =
5
(x−1)(x−2)(x−3)
+
10
(x−2)(x−3)
+
3
(x−3)
B =
5
(x − 1)(x −2)(x −3)
+
10
(x − 2)(x −3)
+
3
(x − 3)
f(x) = 0
f(x) = 0
f(x) = x
4
− 3x − 20
g(x) = x
3
− 2x − 5 (x > 0)
k(x) = x + ln x − 2
h(x) = x

2
− sin πx
f(2) = −10
f(3) = 3
4
− 3 = 52
f(2)f(3) < 0
f

(x) = 4x
3
− 3 > 0 ∀x ∈ (2, 3)
x = x + λf (x) = ϕ(x)
M = max
[2,3]
f

(x) = 105
m = min
[2,3]
f

(x) = 29
q = 1 −
m
M
=
76
105
λ =

−1
M
=
−1
105
⇒ ϕ(x) = x −
x
4
− 3x − 20
105
= −
x
4
105
+
36x
35
+
4
21
ϕ

(x)
| ϕ

(x) | =|
−4x
3
105
+

36x
35
|≤ 1
ϕ(x) ∈ [2, 3] ∀x ∈ [2, 3]
⇒ ϕ(x) = −
x
4
105
+
36x
35
+
4
21
x
∗
x
0
= 2, 2
x
n
= −
1
105
x
4
n−1
+
36
35

x
n−1
+
4
21
| x
n
− x
∗
|=
q
1−q
| x
n
− x
n−1
|=
76
29
| x
n
− x
n−1
|
x
n
| x
n
− x
n−1

|
2, 2
0, 0097 ≤ 0, 01
| x
n
− x
∗
|≤
q
n
1 − q
| x
1
− x
0
|≤ 0, 01 ⇒
(
76
105
)
n
29
105
| 2.23 −2.2 |< 0, 01 ⇒ n > 7, 34
x
∗
= 2, 27 ± 0, 0097
g(2) = −1 g(3) = 16
f(2)f(3) < 0
x = x + λg(x) = ϕ(x)

g

(x) = 3x
2
− 2 > 0 ∀x ∈ [2, 3]
M = max
[2,3]
g

(x) = 25
m = min
[2,3]
g

(x) = 10
λ = −
1
25
q = 1 −
m
M
= 1 −
10
25
=
3
5
⇒ ϕ(x) = x −
1
25

(x
3
− 2x − 5) = −
1
25
x
3
+
27
25
x +
1
5
ϕ

(x) = −
3
25
x
2
+
27
25
| ϕ(x) |< 1 ∀x ∈ [2, 3]
ϕ(x) ∈ [2, 3] ∀x ∈ [2, 3]
x
∗
| x
n
− x

∗
|≤
q
1−q
| x
n
− x
n−1
|= 1.5 | x
n
− x
n−1
|
x
0
= 2 x
n
= −
1
25
(x
n−1
)
3
+
27
25
x
n−1
+

1
5
x
n
| x
n
− x
∗
|
2
2, 04 0, 0266
2, 0636 0, 035
2, 077 0, 02
2, 0848 0, 0116
2, 089 0, 0065 < 0, 01
| x
n
− x
∗
| ≤
q
n
1 − q
| x
1
− x
0
|< 0, 01
⇔ (0, 6)
n

< 0, 1
⇒ n > 5
x
∗
= 2, 089 ±0, 0065
k(1) = −1 k(2) > 0
f(1)f(2) < 0 (1, 2)
k

(x) = 1 +
1
x
x = x + λk(x) = ϕ(x)
M = max
[1,2]
k

(x) = max
[1,2]
(1 +
1
x
) = 2
m = min
[1,2]
k

(x) =
3
2

q = 1 −
m
M
= 1 −
3
4
=
1
4
λ =
−1
M
=
−1
2
⇒ ϕ(x) = x −
x + ln x − 2
2
=
x − ln 2 + 2
2
⇒ ϕ

(x) =
1
2
−
1
2x
ϕ


(x)
| ϕ

(x) |≤ 1 ∀x ∈ [1, 2]ϕ(x) ∈ [
1
2
, 1] ∀x ∈ [1, 2]
x
∗
x
0
= 1, 6
x
n
=
x
n−1
− ln x
n−1
+ 2
2
| x
n
− x
∗
|=
q
1 − q
| x

n
− x
n−1
|=
1
3
| x
n
− x
n−1
|
x
n
1
3
| x
n
− x
n−1
|
1, 6
2.10
−3
< 10
−2
x
∗
= 1, 559 ± 2.10
−3
h(

1
2
) = −
3
4
h(
3
2
) = 1
f(
1
2
)f(
3
2
) < 0 (
1
2
,
3
2
)
h

(x) = 2x − π cos πx
x = x + λh(x) = ϕ(x)
M = max
[
1
2

,
3
2
]
h

(x) = 2 + π ≈ 5, 14
m = min
[
1
2
,
3
2
]
h

(x) = 1
q = 1 −
m
M
=
1 + π
2 + π
λ =
−1
M
=
−1
5, 14

⇒ ϕ(x) = x −
x
2
− sin πx
5, 14
=
−x
2
+ sin πx + 5, 14x
5, 14
⇒ ϕ

(x) =
−2x + π cos πx + 5, 14
5, 14
ϕ

(x)
| ϕ

(x) | =|≤ 1 ∀x ∈ [
1
2
,
3
2
]
ϕ(x) ∈ [
1
2

,
3
2
] ∀x ∈ [
1
2
,
3
2
]
x
∗
x
0
= 0, 8
x
n
=
−x
2
n−1
+sin πx
n−1
+5,14x
n−1
5,14
| x
n
− x
∗

|=
q
1−q
| x
n
− x
n−1
|= 4, 14 | x
n
− x
n−1
|
x
n
4, 14 | x
n
− x
n−1
|
0, 8
8, 69.10
−3
< 10
−2
x
∗
= 0, 7878 ± 8, 69.10
−3
 = 10
−3

g(x) = x
3
− 2x − 5 − 0 (x > 0)
k(x) = x + ln x − 2 = 0
h(x) = x
2
− sin πx = 0
g(2) = −1 g(3) = 16
f(2)f(3) < 0
g

(x) = 3x
2
g

(x) = 6x
M
2
= max
[2,3]
g

(x) = 18
m
1
= min
[2,3]
g

(x) = 10

| x
n
− x
∗
|≤
M
2
2m
1
| x
n
− x
n−1
|
2
=
9
10
| x
n
− x
n−1
|
2
x
0
= 2, 2 f(2, 2)f

(2, 2) > 0 x
0

x
n
=
2x
3
n−1
+ 5
3x
2
n−1
− 2
x
n
9
10
| x
n
− x
n−1
|
2
2, 62.10
−5
< 10
−3
x
∗
= 2, 0946 ± 2, 62.10
−5
k(1) = −1 k(2) > 0

f(2)f(1) < 0
k

(x) = 1 +
1
x
k

(x) = −
1
x
2
M
2
= max
[1,2]
k

(x) = 1
m
1
= min
[1,2]
k

(x) =
3
2
| x
n

− x
∗
|≤
M
2
2m
1
| x
n
− x
n−1
|
2
=
1
3
| x
n
− x
n−1
|
2
x
0
= 1, 5 f(1, 5)f

(1, 5) > 0 x
0
x
n

=
x
n−1
(3 − ln x
n−1
)
x
n−1
+ 1
x
n
1
3
| x
n
− x
n−1
|
2
1, 072.10
−3
3.10
−8
< 10
−3
x
∗
= 1, 557 ± 3.10
−8
h(

1
2
) = −
3
4
h(1) = 1
f(
1
2
)f(1) < 0 (
1
2
, 1)
h

(x) = 2x − π cos πx h

(x) = 2 + π
2
sin πx
(
1
2
, 1)
M
2
= max
[
1
2

,1]
h

(x) ≈ 12
m
1
= min
[
1
2
,1]
h

(x) = 1
| x
n
− x
∗
|≤
M
2
2m
1
| x
n
− x
n−1
|
2
= 6 | x

n
− x
n−1
|
2
x
0
= 0, 8 f(0, 8)f

(0, 8) > 0
x
n
= x
n−1
−
h(x
n−1
)
h

(x
n−1
)
=
x
2
n−1
− πx
n−1
cos πx

n−1
+ sin πx
n−1
2x
n−1
− π cos πx
n−1
x
n
6 | x
n
− x
n−1
|
2
5, 7.10
−3
2, 4.10
−7
< 10
−3
x
∗
= 0, 7872 ± 2, 4.10
−7
√
13
cos 20
0
f(x) = x

2
− 13 = 0
f(3)f(4) < 0
f

(x) = 2x > 0 ∀x ∈ [3, 4]M = max
[3,4]
= 8; m = min
[3,4]
= 6
x = x + λf (x) = ϕ(x)
λ = −
1
M
= −
1
8
q = 1 −
m
M
=
1
4
ϕ(x) = x −
1
8
(x
2
− 13) = −
1

8
x
2
+ x +
13
8
ϕ

(x) = 1 −
1
4
x
| ϕ

(x) | < 1 ∀x ∈ (3, 4)
ϕ(x) ∈ (3, 4) ∀x ∈ (3, 4)
x
∗
| x
n
− x
∗
|≤
q
1 − q
| x
n
− x
n−1
|=

1
3
| x
n
− x
n−1
|
x
0
= 3, 5
x
n
=
−(x
n−1
)
2
+ 8x
n−1
+ 13
8
x
n
| x
n
− x
∗
|
3, 5
3, 59375 0, 03125

3, 6043 0, 00354
3, 6054 3, 75.10
(−4)
√
13 ≈ 3, 6054 ± 3, 75.10
−4
f

(x) = 2x; f

(x) = 2
x
n
= x
n−1
−
f(x
n−1
)
f

(x
n−1
)
=
(x
n−1
)
2
+ 13

2x
n−1
x
0
= 3, 8 f(3, 8)f

(3, 8) > 0 x
0
M
2
= max
[3,4]
f

(x) = 2; m
1
= min
[3,4]
f

(x) = 6
| x
n
− x
∗
|≤
M
2
2m
1

(x
n
− x
n−1
)
2
=
1
6
(x
n
− x
n−1
)
2
x
n
1
6
(x
n
− x
n−1
)
2
3, 8
3, 6105 5, 983.10
−3
3, 60555 4, 12.10
−6

√
13 ≈ 3, 60555 ± 4, 12.10
−6
cos 60
0
= cos(3 × 20
0
) = 4 cos
3
(20
0
) − 3 cos 20
0
=
1
2
cos 20
0
= x (x > 0)
g(x) = 4x
3
− 3x −
1
2
(x > 0)
f(
2
3
)f(1) < 0 (
2

3
, 1)
g

(x) = 12x
2
− 3 > 0 ∀x ∈ (
2
3
, 1)
x = ϕ(x) = x + λg(x)
M = max
[
2
3
,1]
g

(x) = 9
m = min
[
2
3
,1]
g

(x) =
7
3
λ = −

1
M
= −
1
9
q = 1 −
m
M
=
20
27
⇒ ϕ(x) = x −
1
9
(4x
3
− 3x −
1
2
) = −
4
9
x
3
+
12
9
x +
1
18

⇒ ϕ

(x) =
−4x + 4
3
⇒| ϕ

(x) | < 1 ∀x ∈ [
2
3
, 1]
ϕ(x) ∈ [
2
3
, 1] ∀x ∈ [
2
3
, 1]
x
∗
| x
n
− x
∗
|≤
q
1 − q
| x
n
− x

n−1
|=
20
7
| x
n
− x
n−1
|
x
0
= 1
x
n
20
7
| x
n
− x
n−1
|
17
18
4, 28.10
−3
2, 589.10
−4
cos 20
0
≈ 0, 9397 ± 2, 589.10

−4
A =








b =
















−→









−→








⇒ x
1
= 2, x
2
= 1, x
3
= 1
x = (2, 1, 1)
A =









b =
















−→








−→









⇒ x
1
= −1, x
2
= 0, x
3
= 1
x = (−1, 0, 1)
A =















b =




























−→















−→














−→















−→














⇒ x

1
= 1, x
2
= −1, x
3
= −1, x
4
= 0
x = (1 − 1, −1, 0)
x
(
n)  x
(n)
−x
∗

∞
< 10
−2
x
∗
x = (0, 0, 0)
T
A =









b =








q = max(
1
5
,
2
3
,
5
10
) < 1
x
(n)
1
= −
1
5
x
(n−1)
3

+
11
5
x
(n)
2
= −
1
3
x
(n−1)
1
+
1
3
x
(n−1)
3
+
4
3
x
(n)
3
= −
3
10
x
(n−1)
1

−
2
10
x
(n−1)
2
+
6
10
 x
(n)
− x
∗

∞
≤
q
1 − q
 x
(n)
− x
(n−1)

∞
= 2  x
(n)
− x
(n−1)

∞

x
(n)
1
x
(n)
2
x
(n)
3
2  x
(n)
− x
(n−1)

∞
5, 14.10
−3
< 10
−2
5, 14.10
−3
x
(n)
− x
∗

∞
≤
q
n

1 − q
 x
1
− x − 0 
∞
≤ 10
−2
⇔(
2
3
)
n
× 3 × 2, 2 ≤ 10
−2
⇒n ≥ 16
≥ 16
A =








b =









q = max(
1
2
,
2
3
,
1
2
) =
2
3
< 1
x
(n)
1
=
1
2
x
(n−1)
3
−
3
2
x

(n)
2
=
1
3
x
(n−1)
1
−
1
3
x
(n−1)
3
+
2
3
x
(n)
3
= −
1
4
x
(n−1)
1
+
1
4
x

(n−1)
2
+
3
4
 x
(n)
− x
∗

∞
≤
q
1 − q
 x
(n)
− x
(n−1)

∞
= 2  x
(n)
− x
(n−1)

∞
x
(n)
1
x

(n)
2
x
(n)
3
2  x
(n)
− x
(n−1)

∞
−
3
2
2
3
3
4
−
9
8
−
1
12
31
24
−
41
48
−

5
36
97
96
−
191
192
13
288
535
576
3, 42.10
−3
4, 226.10
−3
7, 83.10
−5
8, 3.10
−3
< 10
−2
8, 3.10
−3
x
(n)
− x
∗

∞
≤

q
n
1 − q
 x
1
− x − 0 
∞
≤ 10
−2
⇒n ≥ 15
≥ 15
A =








b =








q = max(

1
5
,
2
3
,
5
10
) =
2
3
< 1
x
(n)
1
= −
1
5
x
(n−1)
3
+
11
5
x
(n)
2
= −
1
3

x
(n)
1
+
1
3
x
(n−1)
3
+
4
3
x
(n)
3
= −
3
10
x
(n)
1
−
2
10
x
(n)
2
+
6
10

x
0
= (1.5, 0.5, 0.5)
 x
(n)
− x
∗

∞
≤
q
1 − q
 x
(n)
− x
(n−1)

∞
= 2  x
(n)
− x
(n−1)

∞
x
(n)
1
x
(n)
2

x
(n)
3
2  x
(n)
− x
(n−1)

∞
8, 94.10
−3
1, 482.10
−3
7, 684.10
−4
7, 684.10
−4
A =








b =









q = max(
1
2
,
2
3
,
1
2
) =
2
3
< 1
x
(n)
1
=
1
2
x
(n−1)
3
−
3
2

x
(n)
2
=
1
3
x
(n)
1
−
1
3
x
(n−1)
3
+
2
3
x
(n)
3
= −
1
4
x
(n)
1
+
1
4

x
(n)
2
+
3
4
 x
(n)
− x
∗

∞
≤
q
1 − q
 x
(n)
− x
(n−1)

∞
= 2  x
(n)
− x
(n−1)

∞
x
0
= (−0, 5, 0, 5, 1, 5)

x
(n)
1
x
(n)
2
x
(n)
3
2  x
(n)
− x
(n−1)

∞
−
3
4
−
1
12
11
12
−
25
24
1
72
73
72

−
143
144
−
1
432
431
432
3, 8594.10
−4
−7, 97.10
−5
2, 8.10
−3
2, 609.10
−4
5.10
−4
< 10
−3
5.10
−4
A =









b =
















−→








−→









⇒ x
1
= 1, x
2
= 3, x
3
= 0
x = (1, 3, 0)
y = e
x
h = 0, 05
y 
2
y 
3
y 
4
y
7, 1.10
−3
4.10
−4
7, 5.10
−3

4.10
−4
7, 9.10
−3
f

(x
0
) ≈
1
h
× (y
0
−

2
y
0
2
+

3
y
0
3
−

4
y
0

4
)
⇒ f

(1) ≈
1
0, 05
× (0, 1394 −
7, 1.10
−3
2
+
4.10
−4
3
) = 2, 71967
f

(1, 05) ≈
1
0, 05
× (0, 1465 −
7, 5.10
−3
2
+
4.10
−4
3
) = 2, 85766

f

(1, 1) ≈
1
0, 05
× (0, 154 −
7, 9.10
−3
2
) = 2, 85766
f

(1, 15) ≈
1
0, 05
(0, 1619) = 3, 238
f

(x
0
) ≈
1
h
2
× (
2
y
0
− 
3

y
0
+
11
12
× 
4
y
0
)
⇒ f

(1) ≈
1
0, 05
2
× (7, 1.10
−3
− 4.10
−4
) = 2, 68
f

(1, 05) ≈
1
0, 05
2
× (7, 5.10
−3
− 4.10

−4
) = 2, 84
f

(1, 1) ≈
1
0, 05
2
× (7, 9.10
−3
) = 3, 16
y = e
−x
h = 0, 05
y 
2
y 
3
y 
4
y
−0, 018
10
−3
−0, 017 −3.10
−4
7.10
−4
5.10
−4

−0, 0163 2.10
−4
9.10
−4
−0, 0154
f

(x
0
) ≈
1
h
× (y
0
−

2
y
0
2
+

3
y
0
3
−

4
y

0
4
)
⇒ f

(1) ≈
1
0, 05
× (−0, 018 −
10
−3
2
−
3.10
−4
3
−
5
4
.10
−4
) = −0, 3725
f

(1, 05) ≈
1
0, 05
× (−0, 017 −
7.10
−4

2
+
2.10
−4
3
) = −0, 34567
f

(1, 1) ≈
1
0, 05
× (−0, 0163 −
9.10
−4
2
) = −0, 335
f

(1, 15) ≈
1
0, 05
(−0, 0154) = −0, 308
f

(x
0
) ≈
1
h
2

× (
2
y
0
− 
3
y
0
+
11
12
× 
4
y
0
)
⇒ f

(1) ≈
1
0, 05
2
× (10
−3
+ 3.10
−4
+
11
12.5.10
−4

) = 0.667
f

(1, 05) ≈
1
0, 05
2
× (7.10
−4
− 2.10
−4
) = 0, 2
f

(1, 1) ≈
1
0, 05
2
× (9.10
−4
) = 0, 36
y = sin x
h = 0, 05
y 
2
y 
3
y 
4
y

−2, 1.10
−3
−1.10
−4
−2, 2.10
−
3 −1.10
−4
−2.10
−4
−2, 4.10
−3