11. 12. 13. 1- misol. (
√
)
binom yoyilmaning o„rta hadini toping.
Yechish: (
√
)
(
√
)
(
√
)
(
)
yoyilmada 19 ta had bor. O„rta had o„ninchi had bo„lib,
(
√
)
(
)
qiymatga teng.
2-misol. (√
√
)
binom yoyilmaning boshidan to„qqizinchi
hadini shu yoyilmaning oxiridan to„qqizinchi hadiga nisbati
songa
teng.
no„malumni toping.
Yechish: Berilgan ifodani binomi formulasi yordamida yozamiz.
(√
√
)
(√
)
(√
)
(
√
)
(√
)
(
√
)
(√
)
(
√
)
(
√
)
yoyilmani boshidan to„qqizinchi hadi
(√
)
(
√
)
ifodaga, oxiridan
to„qqizinchi hadi esa
(√
)
(
√
)
ifodaga teng.
tenglik o„rinli bo„lgani uchun,
tenglik kelib chiqadi. Shartga
ko„ra,
45
(
)
(
)
(
)
(
)
qiymatga teng. U holda bu ifodani soddalashtirib ushbu
(
)
(
)
tenglamani yechamiz.
(
)
(
)
,
,
3-misol. (√
)
binom yoyilmasi beshinchi hadining bi-
nomial koeffitsiyentini uch marta orttirilgani, ikki asosga ko„ra logarifmi
bilan uchinchi hadi binomial koeffitsiyenti ikki asosga ko„ra logarifmi
orasidagi ayirma
ga tengligi ma‟lum bo„lsa, o„zgaruv-
chining qanday qiymatida uchinchi hadni
√ marta orttirilgani bilan
to„rtinchi hadi nisbati 1 ga teng bo„ladi.
Yechish: Masala shartiga ko„ra,
,
u holda
,
ifodani hisoblaymiz, natijada
tenglama hosil bo„ladi, uni soddalashtirib,
ifodani keltirib chiqaramiz va
tenglamani yechamiz, un-
dan
qiymatlar topiladi. Binom ko„rsatkich
bo„lishi mumkin emas. Agar ko„rsatkich
bo„lsa,
√
(√
)
(√
)
tenglama hosil bo„ladi. Bu tenglamani yechamiz. Ifodaning surati
√
(√
)
√
,
46
maxraji esa
(√
)
ko„rinishga ega. Ularning nisbatini hisoblaymiz, natijada ushbu
√
tenglama hosil bo„ladi. Bu tenglamani yechamiz:
(
)
(
)
,
