A=$\left(\frac{x}{x^2-4}+\frac{1}{x+2}+\frac{2}{2-x}\right):\left(1-\frac{x}{x+2}\right)$
=$\left[\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{x+2}-\frac{2}{2-x}\right]:\left(1-\frac{x}{x+2}\right)$
=$\left[\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right]:\left(1-\frac{x}{x+2}\right)$
=$\frac{x+x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}:\frac{x-2-x}{x+2}$
=$\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{2}{x+2}$
=$\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}$
=$\frac{-3}{x-2}$
b) Để $A>0$ thì $\frac{-3}{x-2}>0$
$\Rightarrow x-2<0\\
\Leftrightarrow x<2$
c)Để A nguyên thì $-3⋮x-2$
$\Rightarrow x-2\in Ư\left(-3\right)$
Lập bảng ta tìm được $x\in \left\{\pm 1;3;5\right\}$
