Задание №1

Первое выражение:

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

Второе выражение:

$$\left(\frac{a-8}{a^2-10a+25}-\frac{a}{a^2-25}\right):\frac{a-20}{(a-5)^2}=\left(\frac{a-8}{(a-5)^2}-\frac{a}{(a-5)(a+5)}\right):\frac{a-20}{(a-5)^2}=\frac{(a-8)(a+5)-a(a-5)}{(a-5)^2(a+5)} \cdot \frac{(a-5)^2}{a-20}=\frac{a^2-8a+5a-40-a^2+5a}{(a-5)^2(a+5)} \cdot \frac{(a-5)^2}{a-20}=\frac{2a-40}{(a-5)^2(a+5)} \cdot \frac{(a-5)^2}{a-20}=\frac{2(a-20)}{(a-5)^2(a+5)} \cdot \frac{(a-5)^2}{a-20}=\frac{2}{a+5}$$