Question

Срочно решить ))......

Answer 1

Ответ:

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

Тождество доказано

$5.\;\;(\frac{a-2}{a+2}-\frac{a+2}{a-2}):\frac{2a}{4-a^2} = (\frac{a^2-4a+4-a^2-4a-4}{(a+2)(a-2)}):\frac{2a}{4-a^2} = -\frac{8a}{(a+2)(a-2)}*\frac{(2+a)(2-a)}{2a}=\\\\=\frac{8a}{(2+a)(2-a)}*\frac{(2+a)(2-a)}{2a} = 4$