43
• Agar musbat songa musbat sonni ko‘paytirsak
har doim
musbat chiqadi va ularning o‘rni almashishi natijani
o‘zgartirmaydi (4-qonuniyat, 1-qator).
• Agar manfiy songa musbat son ko‘paytirilsa yoki musbat
songa manfiy son ko‘paytirilsa natija ham manfiy bo‘ladi
(4-qonuniyat, 2-,3-qatorlar).
• Agar manfiy songa manfiy son ko‘paytirilsa natija har
doim musbat bo‘ladi (4-qoniniyat, oxirgi qator).
Yuqoridagi qonuniyatlar
ifoda ichida ketma-ket kelsa,
xuddi guruhlash usulidagiday bir chetdan boshlab hisoblanadi.
Masalan:
( ) ( )
( )
a
b
c a b
c
a b c
− ⋅ − ⋅ − = ⋅ ⋅ − = − ⋅ ⋅
E’tibor bergan bo‘lsangiz, mavzuda manfiy songa musbat
sonni ko‘paytirish haqida yozilmagan. Chunki biz yuqoridagi
qonuniyatdan bilamizki, manfiy songa musbat sonni ko‘pay-
tirsak oxirgi natija ham manfiy bo‘lar ekan:
(
)
a b
a b
− ⋅ = − ⋅
.
Demak manfiy sonni musbat songa ko‘paytirmoqchi
bo‘lsak, oldin ularning qiymatini ko‘paytirib natija oldiga
minus ishorasini qo‘yamiz.
Biz yuqorida ikkita butun sonning ko‘paytmasi
qonuniyat-
larini ko‘rdik. Endi son va qavsli ifoda ko‘paytirilsa yoki
qavsli ifoda bilan qavsli ifoda ko‘paytirlisa ular qanday yo-
yilishi (qavsdan chiqarish yoki qavsni ochish) qanday bo‘li-
shini o‘rganamiz.
Agar qavs oldida bitta son bo‘lsa, qavsni ochganda o‘sha
son qavs ichidagi barcha sonlarga ko‘paytiriladi. Ishorasi
o‘zgarishi yuqoridagi qonuniyat asosida amalga oshiriladi:
(
)
a b c
a b a c
⋅ +
= ⋅ + ⋅
(
)
a b c
a b a c
− ⋅ +
= − ⋅ − ⋅
(
)
a
b c
a b a c
− ⋅ − − = ⋅ + ⋅
(
)
a b c
a b a c
− ⋅ − = − ⋅ + ⋅
Agar qavsli ifoda qavsli ifodaga ko‘paytirilsa,
birinchi
qavs ichidagi har bir son (nechta bo‘lishining ahamiyat yo‘q)
44
keyingi qavs ichidagi barcha sonlarga ko‘paytiriladi. Ishorasi
o‘zgarishi yuqoridagi qonuniyat asosida amalga oshiriladi:
(
) (
)
a b c d
a c a d b c b d
+ ⋅ +
= ⋅ + ⋅ + ⋅ + ⋅
(
) (
)
a b c d
a c a d b c b d
− ⋅ +
= ⋅ + ⋅ − ⋅ − ⋅
(
) (
)
a b c d
a c a d b c b d
+ ⋅ −
= ⋅ − ⋅ + ⋅ − ⋅
(
) (
)
a b c d
a c a d b c b d
− ⋅ −
= ⋅ − ⋅ − ⋅ + ⋅
Yana bir marta eslaymiz, agar ifoda ichida qavslar kelsa
oldin ushbu qavslar ichidagi ifodalar hisoblanadi. Ya’ni
agar qavs ichida yana qavs va uning ichida qo‘shish, ayi-
rish va ko‘paytirish amallari aralashgan ifoda bo‘lsa, de-
mak siz ichki qavs ichidagi ko‘paytirish amallaridan hi-
soblashni boshlaysiz. Keyin ichki qavs ichidagi qo‘shish,
ayirish amallari. Undan bitta oldin turgan qavs ham xuddi
shu tartibda hisoblanadi.
Masalan, quyidagi ifodani hisoblaylik:
(
)
5 8 3 ?
− ⋅ − =
Ushbu ifodani ikkita usulda hisoblash mumkin:
(
)
( )
5 8 3
5 5
5 5
25
− ⋅ − = − ⋅
= − ⋅ = −
(
)
5 8 3
5 8 5 3
40 15
25
− ⋅ − = − ⋅ + ⋅ = − +
= −
Ikkala usulda ham natija bir xil. 1-usulda oldin qavs ichini
hisoblab, keyin ko‘paytma hisoblandi. 2-usulda oldin qavsni
ochib, keyin yig‘indi hisoblandi. Qaysi
usulni tanlash sizning
qo‘lingizda. Agar qaysidir usul ifodani hisoblashni tezlashtirsa
ushbu usuldan foydalanganingiz ma’qul.
Quyidagi ifoda berilgan bo‘lsin:
(
) (
)
9 5 2 4 0 ?
− + ⋅ −
=
45
Ushbu ifoda ham ikkita usulda yechiladi:
(
) (
) ( ) ( )
9 5 2 4 0 6 4
6 4 24
− + ⋅ −
=
⋅
= ⋅ =
(
) (
)
9 5 2 4 0 9 4 9 0 5 4 5 0 2 4 2 0
− + ⋅ −
= ⋅ − ⋅ − ⋅ + ⋅ + ⋅ − ⋅ =
36 0 20 0 8 0 24
− −
+ + − =
Ko‘rib turganingizdek ikkala usulda ham natija bir xil.
Lekin ushbu ifodani 1-usulda ishlash ancha oson va biz
yuqoridagi qavslarni ochish qoidalari to‘g‘ri
ekanligiga ham
i shonch hosil qildik.
Dasturlashda ifodalarni yozishda ko‘paytirish belgisi
o‘rniga * (yulduzcha) qo‘yib yoziladi va qolgan qismi o‘zga-
rishsiz qoladi.
Masalan
(
)
(
)
7 4 5 2 11 20 2 44
⋅ − ⋅ + ⋅
⋅ −
ifodaning
