|
(198
—
202)
:
198.
1) (0,25)
7
· 4
7
;
2)
×
17
17
4
5
5
4
;
3) (
-
0,125)
11
· 8
11
;
4) (
-
0,2)
5
· 5
5
.
199.
1) (
-
0,25)
9
· (
-
4)
9
;
3)
×
3
3
6
11
(8,5) ;
2)
-
× -
7
7
2
7
( 3,5) ;
4)
×
5
5
1
9
(4,5) .
5 — Algebra, 7- sinf
66
200.
1)
×
8
8
5
2 3
6
;
2)
×
5
5
3
4 3
12
;
3)
×
5
5
5
10
2 5
;
4)
×
4
3
3
14
2 7
.
201.
1)
×
×
12
12
12
12
6
4
3
8
;
2)
×
×
10
10
10
10
4
3
2
6
;
3)
×
×
4
4
2
15
3 5 25
;
4)
16
10
4
8
.
202.
1)
×
3
8
8 27
3
;
2)
×
8
2 4
7
2 (7 )
14
;
3)
×
2
5
4
16 3
12
;
4)
×
9
2 5
5 3
2 (2 )
(2 )
.
Kasrni darajaga ko‘taring
(203
—
206):
203.
1)
2
2
3
;
2)
2
5
7
;
3)
2
3
;
a
4)
3
8
.
b
204.
1)
-
2
11
;
m
2)
-
2
13
;
n
3)
-
3
2
;
d
4)
-
3
4
.
c
205.
1)
4
2
;
a
b
2)
4
3
5
;
b
c
3)
æ
ö
ç
÷
è
ø
7
3
2
2
3
;
4)
æ
ö
ç
÷
è
ø
3
2
4
5
7
.
206.
1)
+
3
3
;
a b
2)
+
2
7
2
;
c
3)
+
-
5
;
m n
m n
4)
+
-
7
.
a b
a b
Kasrni daraja shaklida yozing
(207—209)
:
207.
1)
7
7
3
4
;
2)
5
5
2
5
;
3)
3
3
2
;
m
4)
7
7
5
.
a
208
.
1)
6
6
;
x
y
2)
3
3
;
a
b
3)
25
36
;
4)
49
100
.
209.
1)
2
2
(2 )
(3 )
;
b
b
2)
4
4
(4 )
(3 )
;
x
y
3)
-
1
8
;
4)
-
1
27
.
Hisoblang
(210—211):
210.
A
(
x
) nuqta koordinatalar o‘qining qayerida bo‘lishini
chamalab ko‘rsating:
211.
C
(
n
3
) nuqta koordinatalar o‘qining qayerida bo‘lishini
chamalab ko‘rsating:
B(x
3
)
0
C(x
2
)
0
C(x
2
) B(x
3
)
x x x
a
)
b
)
d
)
0
B(x
3
) C(x
2
)
67
212.
1) Yerning massasi 6·10
24
kg ga teng. Quyoshning massasi
2·10
30
kg. Yerning massasi Quyoshning massasidan necha
marta kam?
2) Yerdan Sirius deb nomlanuvchi yulduzgacha bo‘lgan
masofa 83 000 000 000 000 km. Yorug‘lik nuri Yerdan
Siriusgacha necha yilda yetib borishini taqriban hisoblang.
213.
Ifodaning son qiymatini toping:
1)
-
= -
2
, bunda
2;
b
b
b
2
2
2)
-
= -
3
3
, bunda
3.
a
a
a
3
214.
Ifodani daraja shaklida yozing:
1)
-
+
×
+
n
n
n
3
4
2
1
2
5
5
: 5 ;
3)
-
+
-
4 4
1
5
2
;
n
n
n
a
a
a
6
2)
+
-
-
×
n
n
n
4
3
3
2
2
1
3
3
: 3
;
4)
-
+
-
3
2
4
1
n
n
n
b
b
b
5
3
(
n
— natural son).
215
.
n
ning qanday qiymatida tenglik to‘g‘ri bo‘ladi:
1) (4
4
)
n
= 4
12
; 2) (5
n
)
2
= 5
14
; 3) 2
2
n
= 4
5
; 4) 3(3
2
)
n
= 3
11
?
216.
Ko‘paytmani darajaga ko‘taring:
1) (8
a
2
b
4
c
3
)
3
;
2) (9
x
4
y
3
z
7
)
2
;
3) (
-
1,2
x
5
y
7
z
7
)
2
;
4) (
–
1,2
a
3
b
2
c
4
)
5
.
217.
Ifodani asosi
a
bo‘lgan daraja shaklida yozing:
1)
8 5
3 6
;
a a
a a
2)
9 6
5 8
;
a a
a a
3)
3 4
4 3
6 9
( ) ( )
;
a
a
a a
4)
6
3 5
4 2 9
( )
( )
.
a a
a
a
218.
Sonlardan qaysi biri katta:
1) 54
4
mi yoki 21
12
mi;
3) 100
20
mi yoki 9000
10
mi;
2) 10
20
mi yoki 20
10
mi;
4) 6
20
mi yoki 3
40
mi?
219.
To‘g‘ri tenglik hosil qiling. Masala nechta yechimga ega:
1) (...)
2
· (...)
3
=
-
4
a
8
b
9
c
11
; 2) (...)
2
· (...)
3
=
-
8
a
11
b
5
c
7
?
x x x
a
)
b
)
d
)
A(n)
0
B(n
2
)
0
A(n) B(n
2
)
0
B(n
2
) A(n)
68
11-
Teng ko‘paytuvchilar ko‘paytmasini
natural ko‘rsatkichli daraja shaklida yozish
mumkin bo‘lganligi uchun sonning dara-
jasi va sonlar darajalarining ko‘paytmasi ham
birhadlar deyiladi. Masalan, ushbu ifodalar
birhadlar bo‘ladi:
( )
-
-
2
5
2
2
3
1
4
2
,
7 , , 4 ,
.
c
a
a b
220.
Tenglamani yeching:
1)
=
-
1
8
:1,75 7,125 3 ;
x
3)
=
×
18,9 :
0,021 100;
x
2)
+
=
5
1
17
12
18
12
;
x
4)
+
=
754,5 : (37,1
) 15.
x
221.
Sonni standart shaklda yozing:
1) 26 000;
2) 8 647 000;
3) 384 000;
4) Yerdan Quyoshgacha bo‘lgan masofa 149 500 000 km.
Birhad va uning standart shakli
Turli masalalarni yechishda ko‘pincha
2
1
2
,
, 3
ab
abc a b
ko‘ri-
nishdagi algebraik ifodalarga duch kelinadi. Masalan, o‘lcham-
lari 8- rasmda ko‘rsatilgan sovitgichli mashina sig‘imi 3
abc
ga
teng.
3
abc
ifoda birinchisi raqam bilan, qolgan uchtasi
a
,
b
,
c
harflari bilan belgilangan to‘rtta ko‘paytuvchining ko‘paytma-
sidir.
Raqamlar bilan yozilgan ko‘paytuvchilar
sonli ko‘-
paytuvchilar
, harflar bilan belgilangan ko‘paytuvchilar
esa
harfiy ko‘paytuvchilar
deyiladi. Sonli va harfiy
ko‘paytuvchilar ko‘paytmasidan iborat algebraik ifoda
birhad
deyiladi.
Masalan, ushbu ifodalar birhadlardir:
-
×
-
abc
a
ab
a
bab
1
4
, ( 4)
3 ,
( 0,3)
.
c
3a
b
8- rasm.
69
Har bir sonni shu son bilan birning ko‘paytmasi shaklida
yozish mumkin bo‘lgani uchun
a
3
8
, 2,
ko‘rinishdagi ifodalar
ham birhadlar deb hisoblanadi.
Masala.
Birhadning qiymatini hisoblang:
( ) (
)
16
0,5
0,25
ac
a
b
×
×
,
bunda
=
=
=
a
b
c
1
9
3
17
,
34,
.
Harflarning qiymatlarini birhadga qo‘yib, uning qiymati-
ni topamiz, ya’ni yettita sonning ko‘paytmasini hisoblaymiz:
× ×
×
× ×
×
1
9
1
3 17
3
16
0,5
0,25 34.
Sonlarning birinchisini ikkinchisiga, ular qanday yozilgan
bo‘lsa, xuddi shu tartibda ko‘paytirish mumkin:
×
×
×
=
× =
=
×
=
=
×
=
1
16 16 9
48
48
24
;
3
3
3 17
17
17
17
24 1
8
8
1
2
2
;
;
17 3
17 17 4
17 17
16
;
0,5
;
34
4.
Ko‘paytirishning o‘rin almashtirish va guruhlash qonunlarini
qo‘llab, hisoblashni qisqacha bajarish ham mumkin:
( ) (
)
(
)
(
)
=
×
×
×
=
ac
a
b
a a bc
a bc
2
16
0,5
0,25
16 0,5 0,25
2
.
Endi
=
1
3
,
a
=
=
9
17
34,
b
c
bo‘lganda 2
a
2
bc
birhadning qiymatini
topamiz:
×
×
×
×
×
×
=
=
2
1
9
2 34 9
3
17
9 17
2
34
4.
Masalani ikkinchi usul bilan yechishda berilgan birhad an-
cha sodda ko‘rinishda yozilgan edi: 2
a
2
bc
. Bu
–
birhadning
standart shakliga
misol.
Umuman, birinchi o‘rinda turgan faqat bitta son ko‘-
paytuvchidan va har xil asosli harfiy darajalardan
tuzilgan birhadni
standart shakldagi birhad
deyiladi.
70
Har qanday birhadni standart shaklda yozish mumkin.
Buning uchun barcha son ko‘paytuvchilarni o‘zaro
ko‘paytirish va ularning ko‘paytmasini birinchi o‘ringa
yozish kerak. So‘ngra bir xil harfiy ko‘paytuvchilar
ko‘paytmasini daraja shaklida yozish kerak. Harfiy
ko‘paytuvchilar ko‘pincha, shart bo‘lmasa ham, alifbo
tartibida joylashtiriladi.
Birhadning standart shaklida bir xil harflar yo‘qligini
eslatib o‘tamiz.
Standart shaklda yozilgan birhadning son ko‘pay-
tuvchisini shu
birhadning koeffitsiyenti
deyiladi.
Masalan, 2
a
birhadning koeffitsiyenti 2 ga teng;
ab
2
5
6
bir-
hadni ng koeffi tsi yenti
5
6
ga teng, (
-
7)
a
2
b
3
c
birhadning koef-
fitsiyenti (
-
7) ga teng. Oxirgi holda birhadni qavssiz yozish
mumkin:
( )
2 3
2 3
7
7
a b c
a b c
-
= -
.
1 ga teng bo‘lgan koeffitsiyent, odatda, yozilmaydi,
chunki birga ko‘paytirgan bilan son o‘zgarmaydi. Masalan,
1·
abc
2
=
abc
2
, ya’ni
abc
2
birhadning koeffitsiyenti birga teng.
Agar koeffitsiyent (
-
1) ga teng bo‘lsa, bu holda ham birni
va qavslarni yozmasdan, faqat „
-
“ ishorasini qoldirish mum-
kin. Masalan, (
-
1)
abc
=
-
abc
, ya’ni
-
abc
birhadning koef-
fitsiyenti
-
1 ga teng.
So‘z orqali aytilgan fikrni algebraik ifoda yordamida yozing
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