[tex]\it \dfrac{x+8}{x-3} \in \mathbb{Z} \Rightarrow x-3|x+8\ \ \ \ \ (1)\\ \\ \\ Dar,\ x-3|x-3\ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow x-3|x+8-(x-3) \Rightarrow x-3|x+8-x+3 \Rightarrow x-3|11 \Rightarrow\\ \\ \Rightarrow x-3\in\{\pm1,\ \pm11\} \Rightarrow x-3\in \{-11,\ -1,\ 1,\ 11\}|_{+3}\Rightarrow x\in\{-8,\ 2\ ,\ 4,\ 14\}[/tex]
