ĐKXĐ : x + 2 \(\ne0\Rightarrow x\ne-2\)
Ta có A = \(\frac{1-x}{x+2}=\frac{-2-x+3}{x+2}=-1+\frac{3}{x+2}\)
Để A \(\inℤ\Rightarrow\frac{3}{x+2}\inℤ\Rightarrow3⋮x+2\Rightarrow x+2\inƯ\left(3\right)\)
=> \(x+2\in\left\{1;3;-1;-3\right\}\)
=> \(x\in\left\{-1;1;-3;-5\right\}\)