Выражение в дробь, помогите v) (5a^2)//(5ab-b^2)-b//(25a-5b)g) x^2//(x^3-x)+1//(2-2x)

Решите задачу:

$\frac{5a^2}{5ab-b^2} - \frac{b}{25a-5b}= \frac{5a^2}{b(5a-b)} - \frac{b}{5(5a-b)}= \frac{25a^2-b^2}{5b(5a-b)}= \frac{5a+b}{5b}$

$\frac{x^2}{x^3-x}+ \frac{1}{2-2x}= \frac{x^2}{x(x-1)(x+1)}- \frac{1}{2(x-1)}= \frac{2x^2}{2x(x-1)(x+1)}- \frac{x(x+1)}{2x(x-1)(x+1)}= \\\ = \frac{2x^2-x(x+1)}{2x(x-1)(x+1)}= \frac{2x-x-1}{2(x-1)(x+1)}=\frac{x-1}{2(x-1)(x+1)}=\frac{1}{2(x+1)}$