Ta có:
\(\dfrac{x+5}{x-2}=\dfrac{x-2+7}{x-2}=\dfrac{x-2}{x-2}+\dfrac{7}{x-2}=1+\dfrac{7}{x-2}\)
Để \(\dfrac{x+5}{x-2}\) là một số nguyên thì \(\dfrac{7}{x-2}\) phải nguyên
\(\Rightarrow7\) ⋮ \(x-2\)
\(\Rightarrow x-2\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)
\(\Rightarrow x\in\left\{3;1;9;-5\right\}\)