Вопрос в картинках...

Решите задачу:

$( \frac{x}{x^2-4} + \frac{2}{2-x}+ \frac{1}{2+x}):(x-2+ \frac{10-x^2}{2+x} )=\frac{1}{2-x}$

$\boxed{1}\ \ \frac{x}{x^2-4} + \frac{2}{2-x}+ \frac{1}{2+x}=\frac{x}{x^2-4} - \frac{2}{x-2}+ \frac{1}{x+2}= \frac{x-2(x+2)+x-2}{x^2-4}=\\\\ = \frac{x-2x-4+x-2}{x^2-4}=- \frac{6}{x^2-4}= \frac{6}{4-x^2}\\\\ \boxed{2}\ \ x-2+ \frac{10-x^2}{2+x}= \frac{x(2+x)-2(2+x)+10-x^2}{2+x}= \frac{2x+x^2-4-2x+10-x^2}{2+x}= \frac{6}{2+x}\\\\ \boxed{3}\ \ \frac{6}{4-x^2}: \frac{6}{2+x}= \frac{6}{(2-x)(2+x)}\cdot \frac{2+x}{6}= \frac{1}{2-x}$