(x-3) chia hết (x+1) tìm x

$\displaystyle \begin{array}{{>{\displaystyle}l}} ( x-3) \vdots ( x+1)\\ \Leftrightarrow ( x+1) -4\vdots ( x+1)\\ \Leftrightarrow 4\vdots ( x+1) \end{array}$ Vì $\displaystyle x\in \mathbb{Z}$ nên $\displaystyle x+1\in \mathbb{Z}$ $\displaystyle \Rightarrow x+1$ là ước nguyên của 4 $\displaystyle \begin{array}{{>{\displaystyle}l}} \Rightarrow x+1\in \{-1,1,2,-2,4,-4\}\\ \Rightarrow x\in \{-2,0,1,-3,3,-5\} \end{array}$