\(A=\frac{n+3}{n-2}\)
Để A nhận giá trị nguyên thì \(n+3⋮n-2\)
Có: \(n+3=n-2+5\)
Lại có: \(n-2⋮n-2\)
Mà \(n-2+5⋮n-2\)
\(\Rightarrow5⋮n-2\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow n-2\in\left\{\pm1;\pm5\right\}\)
\(\Rightarrow n\in\left\{\pm3;1;7\right\}\)
Vậy \(n\in\left\{\pm3;1;7\right\}\)