Question

Выражение (1/a^2+2ab - 1/a^2-2ab)÷ 4b^2/4b^2-a^2 ^ степень

Answer 1

$(\frac{1}{a^2+2ab}-\frac{1}{a^2-2ab}):\frac{4b^2}{4b^2-a^2}= \frac{1}{ab} \\ \\ 1)\frac{1}{a^2+2ab}-\frac{1}{a^2-2ab}=\frac{1}{a(a+2b)}-\frac{1}{a(a-2b)}=\frac{a-2b-(a+2b)}{a(a-2b)(a+2b)}= \\ \\ =\frac{a-2b-a-2b}{a(a-2b)(a+2b)}= \frac{-4b}{a(a-2b)(a+2b)} \\ \\ 2) \frac{-4b}{a(a-2b)(a+2b)} : \frac{4b^2}{4b^2-a^2}=\frac{-4b}{a(a-2b)(a+2b)} * \frac{4b^2-a^2}{4b^2}= \\ \\ =\frac{4b}{a(a-2b)(a+2b)} * \frac{a^2-4b^2}{4b^2}=\frac{4b(a-2b)(a+2b)}{4ab^2(a-2b)(a+2b)}=\frac{1}{ab}$