\(A=\frac{3n-7}{n+2}=\frac{3.\left(n+2\right)-13}{n+2}=3-\frac{13}{n+2}\inℤ\)
\(\Rightarrow\frac{13}{n+2}\inℤ\)
Mà \(13\inℤ\Rightarrow n+2\inℤ\Rightarrow n\inℤ\)và \(n+2\inƯ\left(13\right)\)
\(\Rightarrow\left(n+2\right)\in\left\{\pm1;\pm13\right\}\)\(\Rightarrow n\in\left\{-1;-3;11;-15\right\}\)