|
147–148
):
147.
1)
3
64
125
;
2)
4
16
81
;
3)
3
3
8
3 ;
4)
5
19
32
7
.
148.
1)
4
4
324 : 4;
2)
3
3
128 : 2000;
3)
3
3
16
2
;
4)
5
5
256
8
;
5)
(
)
-
20
45 : 5;
6)
(
)
-
3
3
3
625
5 : 5.
1
Bu yerda va bundan keyin, agar qo‘shimcha shartlar bo‘lmasa, harflar bilan musbat sonlar
belgilangan deb hisoblaymiz.
58
149.
Ifodani soddalashtiring:
1)
5
5
6 7
2
:
;
a b
ab
2)
4
3
3
81
: 3
;
x y
xy
3)
3
3
2
2
3
9
:
;
x
y
y
x
4)
4
4
3
3
2
8
:
.
b
a
a
b
Hisoblang (
150–151
):
150.
1)
(
)
2
6
3
7
;
2)
( )
3
6
9
;
-
3)
(
)
2
10
32 ;
4)
(
)
4
8
16
.
-
151.
1)
3
729;
2)
1024;
3)
×
9
3
7
3
9
3 ;
4)
×
6
4
5
3
25
5 .
152.
Ifodani soddalashtiring:
1)
( )
6
3
;
x
2)
( )
3
2
3
;
y
3)
(
)
6
3
;
a
b
×
4)
(
)
12
3
4
2
3
;
a
b
×
5)
(
)
6
3
2
;
a b
6)
(
)
4
3 4
3
27
.
a
Hisoblang (
153–155
):
153.
1)
3
3
1
3
4
2
2 ;
×
2)
4
4
3
3
4
4
6 ;
×
3)
4
4
5
2
8
5
15
:
;
4)
3
3
1
2
2
3
22
6 ;
×
5)
(
)
2
3
27
;
6)
(
)
3
3
16
.
154.
1)
×
2
5
3
3
2
;
ab
a b
c
c
2)
×
3
7
5
5
2
3
8
4
;
a
a
b
b
3)
×
4
4
2 2
3 3 2
4
3
;
a b c a b c
abc
4)
×
3
4
3
3
2 2
2
4
2
;
a b
ab
b a b
5)
(
) (
)
5
3
5
3
3
2
;
a
b
×
6)
(
) (
)
4
3
4
3
3 3
2
:
.
a b
ab
155.
1)
3
3
3
49
112
250
;
×
2)
4
4
4
54
120
5
;
×
3)
4
6
3
4
32
2
27
64 ;
+
-
4)
4
3
4
3
1
8
2
3
18 4
256 ;
+
-
5)
-
×
+
3
3
11
57
11
57;
6)
-
×
+
4
4
17
33
17
33.
59
!
Ifodani soddalashtiring (
156–157
):
156.
1)
3
2
3
3
2
4
27 ;
ab
a b
b
×
×
2)
×
×
4
4
3 2
5 2
4
;
abc
a b c
b c
3)
×
5
5
3 2
2 3
5
3
3
;
a b
a b
ab
4)
×
2 5
3
4
4
2
4
8
4
2
.
x y
x y
xy
157.
1)
(
)
3
3 3
3
18
4
;
a
a
+
2)
(
)
(
)
3
8
3
4
2
2
;
x
x
+
3)
(
)
2
3
4 8
3 6
2
;
a b
a b
-
4)
(
)
5
6 12
2
3
5
;
x y
xy
-
5)
(
)
(
)
4
2
8 2
2
8
4
4
;
x y
x y
-
6)
(
)
(
)
5
10
2
5
5
5
:
.
a
a
a a
-
158.
Hisoblang:
1)
3
6
3 9
3
;
×
2)
×
3
4
12
7
343
7
;
3)
(
)(
)
3
3
3
3
3
4
10
25
2
5 ;
-
+
+
4)
(
)(
)
3
3
3
3
3
9
6
4
3
2 .
+
+
-
159.
Isbotlang:
+
-
-
=
4 2 3
4 2 3
2.
12- §.
RATSIONAL KO‘RSATKICHLI DARAJA
1-m a s a l a .
Hisoblang:
4
12
5 .
=
12
3 4
5
(5 ) bo‘lgani uchun
=
=
=
4
12
3 4
3
4
5
(5 )
5
125 .
Shunday qilib,
=
12
4
12
4
5
5 .
Shunga o‘xshash,
-
-
=
15
5
15
5
7
7
ekanligini ko‘rsatish mumkin.
Umuman, agar
n
– natural son,
2
n
³
,
m
– butun son va
m
n
butun son bo‘lsa, u holda
a
> 0 bo‘lganda quyidagi tenglik to‘g‘ri
bo‘ladi:
.
m
n
m
n
a
a
=
(1)
60
Shartga ko‘ra
m
n
— butun son, ya’ni
m
ni
n
ga bo‘lishda
k
butun
son hosil bo‘ladi. Bu holda
m
n
k
=
tenglikdan
m
=
kn
ekanligi kelib
chiqadi. Darajaning va arifmetik ildizning xossalarini qo‘llab, quyidagini
hosil qilamiz:
( )
=
=
=
=
.
m
n
n
n
n m
kn
k
k
n
a
a
a
a
a
Bordi-yu, agar
m
n
butun son bo‘lmasa, u holda
m
n
a
(bunda
a
> 0) daraja (1) formula to‘g‘riligicha qoladigan qilib ta’rif-
lanadi, ya’ni bu holda
m
n m
n
a
a
=
(2)
deb hisoblanadi.
Shunday qilib, (2) formula istalgan butun
m
va istalgan natural
³
2
n
va
a
> 0 son uchun to‘g‘ri bo‘ladi. Masalan,
=
=
=
=
3
4
4
3
12
3
4
16
16
2
2
8;
=
=
× =
5
4
4
5
4
4
4
7
7
7 7
7 7;
-
-
=
=
=
=
=
2
3
3
2
3
3
2
2
3 6
1
1
1
1
9
3
27
3
27
27
.
r
ratsional son — bu
m
n
ko‘rinishidagi son ekanligini, bunda
m
—
butun son,
n
— natural son, ya’ni
m
n
r
=
bo‘lishini eslatib o‘tamiz. Bu
holda (2) formula bo‘yicha
=
=
m
m
n
r
n
a
a
a
ni hosil qilamiz. Shunday
qilib, daraja istalgan ratsional ko‘rsatkich va istalgan musbat asos uchun
aniqlandi. Agar
0
m
n
r
=
>
bo‘lsa, u holda
n
m
a
ifoda faqat
a
> 0
bo‘lgandagina emas, balki
a
= 0 bo‘lganda ham ma’noga ega bo‘ladi.
a
= 0 bo‘lsa,
=
0
0
n
m
. Shuning uchun
r
> 0 bo‘lganda 0
r
= 0 tenglik
o‘rinli deb hisoblanadi.
!
61
(1) va (2) formulalardan foydalanib, ratsional ko‘rsatkichli darajani
ildiz shaklida, va aksincha, tasvirlash mumkin.
(2) formuladan va ildizning xossalaridan
=
mk
m
n
nk
a
a
tenglik kelib chiqishini ta’kidlaymiz, bunda
a
> 0,
m
— butun son
va
n
,
k
— natural sonlar.
Masalan,
3
6
9
4
8
12
7
7
7
=
=
.
Natural ko‘rsatkichli darajaning barcha xossalari istalgan
ratsional ko‘rsatkichli va musbat asosli darajalar uchun to‘g‘ri
bo‘lishini ko‘rsatish mumkin.
Chunonchi, istalgan ratsional
p
va
q
sonlar va istalgan
a
> 0 va
b
> 0 uchun quyidagi tenglik-
lar to‘g‘ri bo‘ladi:
1)
.
4) (
)
,
2)
:
,
5)
.
3) (
)
,
p
q
p q
p
p p
p
p
p
q
p q
p
p q
pq
a
a
b
b
a
a
a
ab
a b
a
a
a
a
a
+
-
æ ö
ç ÷
è ø
×
=
=
=
=
=
Bu xossalar ildizlarning xossalaridan kelib chiqadi. Masalan,
+
×
=
p
q
p q
a
a
a
xossani isbotlaylik.
Aytaylik,
,
m
k
n
l
p
q
=
=
(bunda
n
va
l
— natural sonlar,
m
va
k
—
butun sonlar) bo‘lsin.
+
×
=
m
m k
k
n
n
l
l
a
a
a
(3)
ekanligini isbotlash kerak.
va
m
k
n
l
kasrlarni umumiy maxrajga keltirib, (3) tenglikning chap
qismini
×
=
×
m
ml
kn
k
n
nl
nl
l
a
a
a
a
ko‘rinishida yozamiz.
!
62
Ratsional ko‘rsatkichli darajaning ta’rifidan, ildizning va butun
ko‘rsatkichli darajaning xossalaridan foydalanib, quyidagini hosil
qilamiz:
+
+
+
×
=
×
=
×
=
=
×
=
=
=
.
m
n
ml
kn
k
nl
nl
ml
kn
nl
nl
l
ml kn
m k
nl
nl
ml
kn
ml kn
nl
n
l
a
a
a
a
a
a
a
a
a
a
a
Ratsional ko‘rsatkichli darajaning qolgan xossalari ham shunga
o‘xshash isbot qilinadi.
Darajaning xossalarini qo‘llashga misollar keltiramiz.
1)
+
×
=
=
1
3
1 3
4
4
4 4
7
7
7
7;
2)
2
1
2 1
1
3
6
3 6
2
9 : 9
9
9
9
3;
-
=
=
=
=
3)
( )
9
1
1 9
3
3
3
4
4
4
3
3
3 4
4
4
4
16
16
16
(2 )
2
2
8;
×
×
=
=
=
=
=
=
4)
2
2
2
2
3
3
3
2
3
3
3
3
3
24
(2 3)
2
3
4 3
4 9;
×
=
×
=
×
=
=
5)
æ
ö =
=
=
ç
÷
è
ø
1
1
1
3
3
3
3
1
1
3
3
3
8
2
8
2
27
3
27
3
( )
.
( )
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