|
10- §
.
NATURAL KO‘RSATKICHLI DARAJANING
ARIFMETIK ILDIZI
O‘rta Osiyolik atoqli matematik va astronom
Jamshid ibn Ma’sud
ibn Mahmud G‘iyosiddin al-Koshiy
(taxminan 1430- yilda vafot etgan)
sonlardan istalgan
n
- darajali ildiz chiqarish amalini kashf qildi. Uning
«Arigmetika kaliti» nomli asarining beshinchi bobi «darajaning asosini
aniqlash» deb nomlangan.
53
Quyidagi masalani qaraylik.
1 - m a s a l a .
Tenglamani yeching:
x
4
= 81.
Tenglamani
-
=
4
81 0
x
yoki
-
+
=
2
2
(
9)(
9)
0
x
x
ko‘rinishida yozib
olamiz.
+ ¹
2
9
0
x
bo‘lgani uchun
- =
2
9
0
x
bo‘ladi, bundan,
x
1
= 3,
x
2
= –3.
Shunday qilib,
x
4
= 81 tenglama ikkita haqiqiy ildizga ega:
x
1
= 3,
x
2
= –3. Ularni 81 sonining 4-
darajali ildizlari
, musbat ildizni (3 sonini)
esa 81 sonining 4-
darajali arifmetik ildizi
deyiladi va bunday
belgilanadi:
4
81 . Sunday qilib,
=
4
81
3.
x
n
=
a
tenglama (bunda
n
— natural son,
a
— nomanfiy son) yagona
nomanfiy ildizga ega ekanligini isbotlash mumkin. Bu ildizni
a
sonning
n
-
darajali arifmetik ildizi
deyiladi.
T a ’ r i f .
a nomanfiy sonning n
³
2
natural ko‘rsatkichli
arifmetik ildizi deb, n- darajasi a ga teng bo‘lgan nomanfiy
sonni aytiladi.
a
sonning
n
- darajali arifmetik ildizi bunday belgilanadi:
n
a
.
a
son
ildiz ostidagi ifoda
deyiladi. Agar
n
= 2 bo‘lsa, u holda
2
a
o‘rniga
a
yoziladi.
Ikkinchi darajali arifmetik ildiz
kvadrat ildiz
ham deyiladi,
3- darajali ildiz esa
kub ildiz
deyiladi.
So‘z
n
- darajali arifmetik ildiz haqida yuritilayotgani aniq bo‘lgan
hollarda qisqacha «
n
- darajali ildiz» deyiladi.
Ta’rifdan foydalanib,
n
a
ning
b
ga tengligini isbotlash
uchun:
1)
0; 2)
n
b
b
a
³
=
ekanligini ko‘rsatish kerak.
Masalan,
3
64
4
=
, chunki 4 > 0 va 4
3
= 64.
Arifmetik ildizning ta’rifidan, agar
a
³
0 bo‘lsa, u holda
( )
,
n
n
n
n
a
a
a
a
=
=
bo‘lishi kelib chiqadi.
Masalan,
( )
5
6
6
5
7
7,
13
13.
=
=
n-
darajali ildiz izlanayotgan amal
n- darajali ildiz chiqarish amali
deyiladi. U
n
- darajaga ko‘tarish amaliga teskari amaldir.
!
!
54
2- m a s a l a .
x
3
= –8 tenglamani yeching.
Bu tenglamani –
x
3
= 8 yoki (–
x
)
3
= 8 kabi yozish mumkin. –
x
=
y
deb belgilaymiz, u holda
y
3
= 8 bo‘ladi.
Bu tenglama bitta ildizga ega:
=
=
3
8
2.
y
y
3
= 8 tenglama manfiy
ildizga ega emas, chunki
y
< 0 bo‘lganda
y
3
< 0 bo‘ladi.
y
= 0 soni ham
bu tenglamaning ildizi bo‘la olmaydi.
Shunday qilib,
y
3
= 8 tenglama faqat bitta
y
= 2 ildizga ega, demak,
x
3
= –8 tenglama ham faqat bitta ildizga ega:
x
= –
y
= –2.
J a v o b :
x
= –2.
x
3
= –8 tenglamaning yechimini qisqacha bunday yozish mumkin:
= -
= -
3
8
2.
x
Umuman, istalgan toq 2
k
+ 1 natural son uchun
a
< 0
bo‘lganda
x
2
k
+1
=
a
tenglama faqat bitta, buning ustiga manfiy
ildizga ega. Bu ildiz xuddi arifmetik ildiz kabi bunday belgi-
lanadi:
2
1
k
a
+
. Uni
manfiy sonning toq darajali ildizi
deyiladi.
Masalan,
-
= -
-
= -
3
5
27
3,
32
2 .
Manfiy
a
sonning toq darajali ildizi bilan
- =
a
a
sonning arifmetik
ildizi orasida ushbu tenglik mavjud:
k
k
k
a
a
a
+
+
+
= -
- = -
2
1
2
1
2
1
.
Masalan,
-
= -
= -
5
5
243
243
3.
M a s h q l a r
131.
(Og‘zaki.) 1) Sonning arifmetik kvadrat ildizini toping:
1
289
1; 0; 16; 0,81; 169;
.
2) Sonning arifmetik kub ildizini toping:
1
27
1; 0; 125;
; 0,027; 0,064.
3) Sonning to‘rtinchi darajali arifmetik ildizini toping:
16
256
81
625
0; 1; 16;
;
; 0,0016.
!
55
Hisoblang (
132–134
):
132.
1)
6
3
36 ;
2)
12
2
64 ;
3)
( )
2
4
1
25
; 4)
8
4
225 .
133.
1)
3
6
10 ;
2)
3
12
3 ;
3)
( )
12
4
1
2
;
4)
( )
16
4
1
3
.
134.
1)
-
3
8; 2)
-
15
1; 3)
-
3
1
27
; 4)
-
5
1024; 5)
-
3
3
34 ; 6)
-
7
7
8 .
135.
Tenglamani yeching:
1)
x
4
= 81;
2)
5
1
32
;
x
= -
3)
= -
5
5
160;
x
4)
=
6
2
128.
x
136.
x
ning qanday qiymatlarida ifoda ma’noga ega bo‘ladi:
1)
-
6
2
3;
x
2)
+
3
3;
x
3)
- -
3
2
2
1;
x
x
4)
4
2 3
2
4
?
x
x
-
-
Hisoblang (
137–138
):
137.
1)
-
+
3
6
1
8
125
64;
2)
5
3
32
0,5
216;
-
-
3)
-
+
4
4
1
3
81
625;
4)
3
4
1
4
1000
256;
-
-
5)
-
+
4
5
1
32
0, 0001 2 0,25
;
6)
+ -
-
3
4
5
1
243
0,001
0,0016.
138.
1)
+
×
-
9
17
9
17;
2)
(
)
+
-
-
2
3
5
3
5 ;
3)
(
)
+
+
-
2
5
21
5
21 ;
4)
+
-
-
+
-
3
2
3
2
3
2
3
2
.
139.
1) a)
³
<
2; b)
2
x
x
bo‘lganda
-
3
3
(
2)
x
ni soddalashtiring;
2)
£
>
a)
3; b)
3
x
x
bo‘lganda
-
6
(3
)
x
ni soddalashtiring.
140.
<
<
1987
1988
n
bo‘ladigan nechta natural son
n
bor?
56
11- §.
ARIFMETIK ILDIZNING XOSSALARI
n
- darajali arifmetik ildiz quyidagi xossalarga ega:
Agar
0,
0,
a
b
³
>
n
,
m
natural sonlar bo‘lib,
2,
2
n
m
³
³
bo‘lsa, u holda quyidagi tengliklar to‘g‘ri bo‘ladi:
1.
.
n
n
n
ab
a b
=
3.
( )
.
m
n
m
n
a
a
=
2.
.
n
n
n
a
a
b
b
=
4.
.
m n
nm
a
a
=
1- xossada
b
son 0 ga teng bo‘lishi ham mumkin. 3- xossada
m
son,
agar
a
> 0 bo‘lsa, istalgan butun son bo‘lishi mumkin.
Masalan,
=
.
n
n
n
ab
a b
ekanligini isbot qilamiz.
Arifmetik ildizning ta’rifidan foydalanamiz:
1)
³
0,
n
n
a b
chunki
³
³
0 va
0
a
b
.
2)
(
)
=
,
n
n
n
a b
ab
chunki
(
)
( ) ( )
=
=
n
n
n
n
n
n
n
a b
a
b
ab
.
Qolgan xossalar ham shunga o‘xshash isbot qilinadi.
Arifmetik ildizning xossalarini qo‘llashga misollar keltiramiz.
1)
×
=
× =
=
=
4
4
4
4
4
4
27
3
27 3
81
3
3.
2)
=
=
=
3
3
3
3
3
64
256
4
256 4
4
625
5
625 5
5
125
:
:
;
3)
=
=
=
7
7
21
7 3
3
5
( 5 )
5
125;
4)
=
=
=
12
3
12
4
12
4096
4096
2
2;
5)
( )
-
-
=
=
=
2
4
2
4
4
1
1
81
3
9
9
.
M a s a l a .
Ifodani soddalashtiring:
(
)
4
4
3 2
3
12 6
,
a b
a b
bunda
a
> 0,
b
> 0.
!
57
Arifmetik ildizning xossalaridan foydalanib, hosil qilamiz:
(
)
4
4
3 2
3 2
3 2
2
6
12 6
3
12 6
.
a b
a b
a b
a b
a b
a b
ab
=
=
=
M a s h q l a r
1
Hisoblang (
141–144
):
141.
1)
3
343 0,125;
×
2)
×
3
864 216;
3)
×
4
256 0, 0081;
4)
×
5
32 100000.
142.
1)
×
3
3
3
5 7 ;
2)
×
4
4
4
11 3 ;
3)
( )
×
5
5
5
0,2
8 ; 4)
( )
7
7
7
1
3
21 .
×
143.
1)
×
3
3
2
500;
2)
×
3
3
0,2
0,04; 3)
×
4
4
324
4;
4)
×
5
5
2
16.
144.
1)
×
5
10
15
3
2 ;
2)
×
3
3
6
2 5 ;
3)
( )
8
12
4
1
3
3
;
×
4)
( )
20
30
10
1
2
4
.
×
145.
Ildiz chiqaring:
1)
3
3 6
64
;
x z
2)
4
8 12
;
a b
3)
10
20
5
32
;
x y
4)
6
12 18
.
a b
146.
Ifodani soddalashtiring:
1)
×
3
3
2
2
2
4
;
ab
a b
2)
×
4
4
2 3
2
3
27
;
a b
a b
3)
×
3
4
4
;
ab
a c
c
b
4)
3
3
2
16
1
2
.
a
ab
b
×
Hisoblang (
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