I I I B O B .
RATSIONAL
KO‘RSATKICHLI DARAJA
9- §.
BUTUN KO‘RSATKICHLI DARAJA
Natural ko‘rsatkichli darajaning xossalari qaralganda darajalarni
bo‘lishning
a
n
:
a
m
=
a
n
-
m
(1)
xossasi
n
>
m
va
a
¹
0 bo‘lganda to‘g‘riligi ta’kidlangan edi.
Agar
n
£
m
bo‘lsa, u holda (1) tenglikning o‘ng qismidagi
n – m
daraja ko‘rsatkich manfiy son yoki nolga teng bo‘ladi.
Manfiy va nol ko‘rsatkichli daraja
shunday aniqlanadiki, (1) tenglik
faqat
n > m
bo‘lgandagina emas, balki
n
£
m
bo‘lganda ham to‘g‘ri bo‘ladi.
Masalan,
n
= 2,
m
= 5 bo‘lganda (1) formula bo‘yicha quyidagini hosil
qilamiz:
-
-
=
=
2
5
2 5
3
:
.
a
a
a
a
Ikkinchi tomondan,
2
2
2
5
2 3
3
5
1
:
.
a
a a
a
a
a
a
a
=
=
=
Shuning uchun
-
=
3
3
1
a
a
deb hisoblanadi.
1 - t a ’ r i f .
Agar a
¹
0
va n – natural son bo‘lsa, u holda
1
n
n
a
a
-
=
bo‘ladi.
!
r
V
= 100 sm
3
a
a
a
V
= 100 sm
3
a < r
a > r
?
49
M i s o l l a r :
1)
-
=
=
3
3
1
1
8
2
2
;
2)
4
4
1
1
81
( 3)
( 3)
;
-
-
-
=
=
3)
3
3
1
1
0,125
( 0,5)
( 0,5)
8.
-
-
-
=
= -
= -
Agar
n
=
m
bo‘lsa, u holda (1) formula bo‘yicha quyidagini hosil
qilamiz:
-
=
=
0
:
.
n
n
n n
a
a
a
a
Ikkinchi tomondan,
=
=
:
1.
n
n
n
n
a
a
a
a
Shuning uchun
a
0
= 1 deb
hisoblanadi.
2 - t a ’ r i f .
Agar a
¹
0
bo‘lsa, u holda a
0
= 1
bo‘ladi.
Masalan, 3
0
= 1,
( )
0
2
5
1.
=
Manfiy ko‘rsatkichli darajalardan sonni
standart shaklda yozishda
foydalanilgan. Masalan,
4
4
1
10
0,00027
2,7
2,7 10 .
-
=
×
=
×
Natural ko‘rsatkichli darajalarning barcha xossalari istalgan butun
ko‘rsatkichli darajalar uchun ham to‘g‘ri bo‘ladi.
Istalgan
a
¹
0,
b
¹
0 va istalgan butun
n
va
m
lar uchun
quyidagi tengliklar to‘g‘ri:
1.
a
n
a
m
=
a
n
+
m
.
4.
a
n
:
a
m
=
a
n
–
m
.
2. (
a
n
)
m
=
a
nm
.
5. (
ab
)
n
=
a
n
b
n
.
3.
( )
=
.
n
n
n
a
a
b
b
Masalan,
n
< 0 bo‘lganda (
ab
)
n
=
a
n
b
n
tenglikning to‘g‘riligini isbot
qilamiz.
n
– butun manfiy son bo‘lsin. U holda
n
= –
k
(bunda
k
– natural
son). Manfiy ko‘rsatkichli darajaning ta’rifidan va natural ko‘rsatkichli
darajaning xossalaridan foydalanib, quyidagini hosil qilamiz:
!
!
4 – Algebra, 9- sinf uchun
50
-
-
-
=
=
=
=
×
=
=
×
=
1
1
1
1
(
)
( )
( )
.
n
k
k
k k
k
k
k
k
n n
ab
a b
a
b
ab
ab
a
b
a b
Butun ko‘rsatkichli darajalarning boshqa xossalari ham shunga
o‘xshash isbot qilinadi.
Butun ko‘rsatkichli darajalarning xossalarini qo‘llashga misollar
keltiramiz:
3
11
6
3 11 6
2
2
3
3 ( 2)
2 6
6 4
2
2
2 ( 2)
4
3
3
3
1) 4
4
4
4
4
16;
2)
9
.
p
p
p
q
q
q
p q
-
-
- + -
-
-
- × -
-
× -
-
×
×
×
=
=
=
æ
ö =
=
=
ç
÷
è
ø
M a s a l a .
6
2
4
2
3
1
(
)(
)
a a
a
a
a
-
-
-
-
+
ifodani soddalashtiring:
(
)
2
4
1
1
1
6
2
4
2
3
1
6
2
3
(
)(
)
a
a
a
a
a a
a
a
a
a
+
-
-
-
-
+
=
-
×
=
-
+
=
×
×
= -
2
6
4
2
1
1
(1
)
1.
a
a
a
a
a
a
M a s h q l a r
109.
Hisoblang:
1)
+ -
- -
+ -
3
3
2
5
2
( 3)
( 2)
( 1) ;
2)
-
- -
-
2
3
4
( 7)
( 4)
3 ;
3)
×
- ×
+
3
3
3
13 2
9 2
2 ;
4)
-
- -
- -
3
3
3
6( 2)
5( 2)
( 2) .
110.
Ifodani natural ko‘rsatkichli daraja shaklida tasvirlang:
1)
×
2
15
13
7 7
7
;
2)
×
×
×
3
10
4
15
5 5 5
5 5
;
3)
2 8 3
9 2
;
a a b
a b
4)
3 5 9
10 7
.
c d c
c d
111.
(Og‘zaki.) Hisoblang:
1) 1
-
5
;
2) 4
-
3
;
3) (
-
10)
0
; 4) (
-
5)
-
2
; 5)
( )
4
1
2
;
-
6)
( )
1
3
4
-
.
112.
Manfiy ko‘rsatkichli daraja shaklida yozing:
1)
5
1
4
;
2)
3
1
21
;
3)
7
1
;
x
4)
9
1
.
a
51
Hisoblang (
113–114
):
113.
1)
( )
3
10
3
;
-
2)
( )
2
9
11
;
-
-
3) (0,2)
-
4
;
4) (0,5)
-
5
;
5)
-
(
-
17)
-
1
;
6)
-
(
-
13)
-
2
.
114.
1) 3
-
1
+ (
-
2)
-
2
;
2)
( )
3
2
2
3
4 ;
-
-
-
3)
-
-
+
2
5
(0,2)
(0,5) ;
4)
-
-
-
- -
3
3
( 0,1)
( 0,2) .
115.
(Og‘zaki.) Bir bilan taqqoslang:
1) 12
-
3
;
2) 21
0
;
3) (0,6)
-
5
;
4)
( )
4
5
19
.
-
116.
Ifodani manfiy ko‘rsatkichsiz daraja shaklida yozing:
1)
-
-
2
(
) ;
x y
2)
-
+
3
(
) ;
x y
3)
5 8
3
;
c
-
4)
-
3
4
9
;
a b
5)
-
-
1 2
3
;
a b c
6)
-
-
2
1
4
.
a b c
Hisoblang (
117–118
):
117.
1)
( ) ( )
3
1
1
7
7
;
-
×
2)
( ) ( )
4
1
1
5
5
;
-
-
× -
3)
-
×
7
10
0,3 0,3 ; 4)
-
×
×
5
3
17
17 17.
118.
1)
7
10
9 : 9 ; 2)
-
2
2
(0,2) : (0,2) ; 3)
( ) ( )
12
10
2
2
13
13
:
;
-
4)
( ) ( )
3
1
2
2
5
5
:
.
-
119.
Darajani darajaga ko‘taring:
1)
-
3
5
( ) ;
a
2)
-
-
2
4
(
) ;
b
3)
-
3 7
(
) ;
a
4)
7
4
( ) .
b
-
120.
Ko‘paytmani darajaga ko‘taring:
1)
-
2 3
(
) ;
ab
2)
-
2
1 4
(
) ;
a b
3)
-
2
6
(2 ) ;
a
4)
-
3
4
(3 ) .
a
121.
Amallarni bajaring:
1)
2
8
7
;
a
b
-
æ
ö
ç
÷
è
ø
2)
4
5
3
;
m
n
-
-
-
æ
ö
ç
÷
è
ø
3)
2
6
4
2
3
;
x
y
-
æ
ö
ç
÷
è
ø
4)
3
5
3
4
.
x y
z
-
-
æ
ö
ç
÷
è
ø
122.
1)
=
=
5,
6,7
x
y
bo‘lganda,
-
-
-
æ ö
-
× ç ÷
è ø
2
2
2
2
1
(
4
)
y
x y
y
ning qiymatini
hisoblang; 2)
=
2,
a
= -
3
b
bo‘lganda (
4
4
2
1 4
0 4
2
(
)
) :
a b
b
a b
a b
-
-
-
ning
qiymatini hisoblang.
52
Standart shaklda yozing (
123–124
):
123.
1) 200 000
4
;
2) 0,003
3
;
3) 4000
-
2
;
4) 0,002
-
3
.
124.
1) 0,0000087;
2) 0,00000005086;
3)
1
125
;
4)
1
625
.
125.
Oynani silliqlash jarayoni uning sirtidagi o‘yiqliklar chuqurligi
3
∙
10
-
3
mm dan ortmaydigan bo‘lganda to‘xtatiladi. Shu sonni
o‘nli kasr shaklida yozing.
126.
O‘rta og‘irlikdagi vodorod 0,00 000 000 001 sekundgina «yashaydi»
(mavjud bo‘ladi). Shu sonni manfiy ko‘rsatkichli daraja shaklida
yozing.
127.
Gripp virusining o‘lchamlari taqriban 10
-
4
mm ni tashkil qiladi.
Shu sonni o‘nli kasr shaklida yozing.
128.
Kasrni daraja shaklida tasvirlang va uning qiymatini
a
ning
berilgan qiymatida toping:
1)
8
7
2
,
0,8;
a a
a
a
-
-
=
2)
15 3
13
1
2
,
.
a a
a
a
=
129.
Hisoblang:
1)
-
-
-
-
-
+
7
7
6 8
2
(( 20) ) : (( 20) )
2 ;
2)
( )
2
4
6
13
2
1
17
(( 17) ) : (( 17)
)
.
-
-
-
-
-
-
-
130.
Soddalashtiring:
1)
-
-
-
-
-
-
-
-
-
-
+
×
-
×
-
+
3
3
2
2
1
2
1
1
2
1
(
) (
)
(
) ;
a
b
a
b
a
a b
b
2)
-
-
-
-
-
-
-
-
×
+
+
2
2
2
1
1
2
1
(
) (
) .
a b ab
a
a b
b
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