☰ Reja · Matematika

Daraja va ildizlar

6-hafta · blok 1.4 · 3 dars
3

Ildizlarni taqqoslash va maxrajni irratsionallikdan qutqarish

maktab darsligi: algebra 8-sinf, ildiz ostidan ko‘paytuvchi chiqarish; algebra 9-sinf
Maqsad: Ildizlarni taqqoslash va kasr maxrajidagi ildizdan qutulish.
Qoida va g‘oya

Ildizli sonlarni taqqoslash uchun ularni bir xil shaklga keltiring: koeffitsiyentni ildiz ostiga kiriting. Masalan, 3 sqrt(2) = sqrt(18), 2 sqrt(5) = sqrt(20), demak 3 sqrt(2) < 2 sqrt(5). Musbat sonlarda kattaroq ildiz osti kattaroq songa mos. Maxrajni irratsionallikdan qutqarish: a / sqrt(b) = a sqrt(b) / b. Maxraj ikki had yig‘indisi bo‘lsa, qo‘shma ifodaga ko‘paytiring va kvadratlar ayirmasidan foydalaning: 1 / (sqrt(a) − sqrt(b)) = (sqrt(a) + sqrt(b)) / (a − b). Bu usul ko‘p testlarda uchraydi.

Namunaviy masalalar
3 sqrt(2) va 2 sqrt(5) ni taqqoslang.
Yechim: 3 sqrt(2) = sqrt(18), 2 sqrt(5) = sqrt(20). 18 < 20.
Javob: 3 sqrt(2) < 2 sqrt(5)
6 / sqrt(3) ning maxrajini irratsionallikdan qutqaring.
Yechim: 6 sqrt(3) / 3 = 2 sqrt(3).
Javob: 2 sqrt(3)
1 / (sqrt(3) − 1) ni soddalashtiring.
Yechim: Qo‘shma ifodaga ko‘paytiramiz: (sqrt(3) + 1) / (3 − 1) = (sqrt(3) + 1) / 2.
Javob: (sqrt(3) + 1) / 2
Tipik xatolar
  • Ildizlarni taqqoslashda koeffitsiyentni ildiz ostiga kiritishni unutish: 3 sqrt(2) = sqrt(18), sqrt(6) emas.
  • Maxrajdagi ayirmani (sqrt(a) − sqrt(b)) yig‘indiga ko‘paytirishda ishorani adashtirish.
  • 1 / (sqrt(3) − 1) ni sqrt(3) − 1 ga ko‘paytirish: qo‘shma ifoda ishorasi qarama-qarshi bo‘lishi kerak.
Mashq
1
10 / sqrt(5) ning maxrajini irratsionallikdan qutqaring.
2
(sqrt(7) − 2)(sqrt(7) + 2) ni hisoblang.
3
3 sqrt(2) va 4 sonlarini taqqoslang.
4
4 / (sqrt(6) − sqrt(2)) ni soddalashtiring.