a) x$\neq$ ±2
A=$\frac{4x}{x²-4}$ +$\frac{2}{2-x}$ +$\frac{1}{x+2}$
=$\frac{4x}{x²-4}$ -$\frac{2}{x-2}$ +$\frac{1}{x+2}$
=$\frac{4x-2(x+2)+x-2}{(x+2)(x-2)}$
=$\frac{4x-2x-4+x-2}{(x+2)(x-2)}$
=$\frac{3x-6}{(x+2)(x-2)}$
=$\frac{3(x-2)}{(x+2)(x-2)}$
=$\frac{3}{x+2}$
b) x$\neq$ ±2
A=$\frac{3}{x+2}$
Để A nguyên thì $\frac{3}{x+2}$ nguyên
⇒x+2 ∈$Ư(3)=\{-1;1;-3;3\} $
⇔x ∈ {-3;-1;-5;1}