Bài 2.
\(A=\dfrac{x-3}{x+1}=\dfrac{x+1-4}{x+1}=\dfrac{x+1}{x+1}-\dfrac{4}{x+1}=1-\dfrac{4}{x+1}\)
Để A nguyên thì \(\dfrac{4}{x+1}\in Z\Rightarrow x+1\in U\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
x+1=1 => x=0
x+1=-1 => x=-2
x+1=2 => x= 1
x+1=-2 => x=-3
x+1=4 => x=3
x+1=-4 => x=-5
Vậy \(x\in\left\{0;-2;1;-3;3;-5\right\}\) thì A nguyên