...
...
...
...
...
110
,
2
1
2
1
1
2
1
1
3
x
x
x
x
,
...
1
1
1
0
2
n
n
x
x
x
x
,
2
1
...
2
1
...
2
2
1
2
1
1
0
2
n
n
n
n
n
n
n
x
x
x
x
x
0
1
0
0
2
1
1
2
1
2
n
n
n
n
n
n
n
n
n
n
x
x
x
x
f
.
7.
x
e
x
x
f
1
Javob:
x
x
n
n
e
x
x
f
n
n
n
x
,
!
1
1
1
1
2
1
.
8.
x
x
x
f
1
1
ln
Javob:
1
1
...,
1
2
2
...
3
2
2
1
1
ln
1
2
3
x
n
x
x
x
x
x
x
f
n
9.
2
2
1
x
x
x
f
Javob:
,
!
!
2
!
!
1
2
...
2
1
1
2
2
2
2
2
2
n
x
n
n
x
x
x
x
x
f
1
1
x
10.
x
x
x
f
1
1
ln
Javob:
3
2
3
1
2
1
1
2
1
1
1
1
ln
x
x
x
x
x
x
f
1
1
,
...
4
1
3
1
2
1
1
4
x
x
11.
2
2
3
1
ln
x
x
x
f
Javob:
,
2
1
1
2
3
1
ln
1
2
n
x
x
x
x
f
n
n
n
2
1
2
1
x
12. Binomial qator yordami bilan
1
x
bo’lganda
...
6
4
2
5
3
1
4
2
3
1
2
1
1
1
1
6
4
2
2
x
x
x
x
111
...
2
...
4
2
1
2
...
3
1
2
n
x
n
n
ekanini ko’rsating va qatorni hadma-had integrallab,
x
arcsin
uchun qatorga yoying.
Javob:
...
1
2
2
...
4
2
1
2
...
3
1
...
5
4
2
3
1
3
2
1
arcsin
1
2
5
3
n
x
n
n
x
x
x
x
n
.
13. Binomial qator yordami bilan
1
x
bo’lganda
...
!
3
2
5
3
1
!
2
2
3
1
2
1
1
1
1
6
3
4
2
2
2
x
x
x
x
ekanligini ko’rsating va qatorni hadma-had integrallab,
2
1
ln
x
x
funksiya uchun qatorga yoying.
4-§. Qatorlarning taqribiy hisoblashlarga tatbiqi
1-misol.
4
10
ни
1
,
1
ln
gacha aniqlikda hisoblang.
Yechish. Ma’lumki
.
.
.
1
.
.
.
4
3
2
1
ln
1
4
3
2
n
x
x
x
x
x
x
n
n
1
1
x
da
1
,
0
x
deb olsak
..
.
.
4
1
,
0
3
1
,
0
2
1
,
0
1
,
0
1
,
1
ln
4
3
2
Bu ishoralari almashinuvchi Leybnis qatori. Qator to’rtinchi
hadining absolyut qiymati
4
10
dan kichik bo’lgani sababli,
4
10
ни
1
,
1
ln
gacha aniqlikda hisoblash uchun qatorning
uchta
3
n
hadini olish yetarlidir. Demak,
0953
,
0
3
001
,
0
2
01
,
0
1
,
0
1
,
1
ln
.
112
0953
,
0
1
,
1
ln
.
2-misol.
5
10
ни
2
ln
gacha aniqlikda hisoblang.
Yechish. Logarifmlarni taqribiy hisoblashda
3
1
2
3
1
1
2
1
2
ln
1
ln
N
N
N
N
...
1
2
5
1
5
N
dan foydalanamiz. Bu yerda
1
N
bo’ladi.
...
3
1
2
1
...
3
5
1
3
3
1
3
1
2
2
ln
1
2
5
3
n
n
.
1
n
R
qoldiq hadni baholaymiz.
...
3
1
5
2
1
3
1
3
2
1
2
5
2
1
2
1
n
n
n
n
n
R
8
3
3
2
9
2
3
1
3
1
1
3
3
2
2
3
2
2
3
2
n
n
n
n
n
5
1
2
10
3
3
2
4
1
n
n
.
Bundan
5
1
2
10
3
3
2
4
n
n
va bu tengsizlik
4
n
bo’lganda o’rinli bo’ladi. Demak,
69314
,
0
3
9
1
3
7
1
3
5
1
3
3
1
3
1
2
2
ln
9
7
5
3
yoki
69314
,
0
2
ln
.
3-misol.
3
1
,
0
10
ни
e
gacha aniqlikda hisoblang.
Yechish.
x
e
ning yoyilmasidan foydalanamiz.
n
n
x
R
x
n
x
x
e
1
2
!
1
1
.
.
.
!
2
1
1
,
113
x
x
n
R
n
0
,
!
1
.
1
,
0
x
desak,
3
1
,
0
e
e
e
bo’ladi.
U holda
001
,
0
!
10
3
n
R
n
n
tengsizlik
3
n
bo’lganda
o’rinli bo’ladi. Demak,
3
1
,
0
10
ни
e
gacha aniqlikda
hisoblash uchun qatorning uchta hadini hisoblash kifoya,
ya’ni
105
,
1
200
1
10
1
1
1
,
0
e
yoki
105
,
1
1
,
0
e
.
4-misol.
0001
,
0
ни
130
3
gacha aniqlikda taqribiy
hisoblang.
Yechish. Ma’lumki
3
2
!
3
2
1
!
2
1
1
1
x
m
m
m
x
m
m
mx
x
m
1
1
,
.
.
.
!
1
...
1
...
x
x
n
n
m
m
m
n
3
130
ni quyidagi ko’rinishda yozamiz
3
1
3
3
3
25
1
1
5
25
1
1
125
5
125
130
.
3
1
m
bo’lsa
...
!
3
2
3
1
1
3
1
3
1
!
2
1
3
1
3
1
3
1
1
1
3
2
3
1
x
x
x
x
...
!
4
3
8
5
2
1
!
3
3
5
2
1
!
2
3
2
1
3
1
1
4
4
3
3
2
2
x
x
x
x
.
Endi hosil bo’lgan qatorda
x
ning o’rniga
25
1
ni
qo’yamiz.