Để \(n\in Z\)thì \(A\in Z\)
\(A=\frac{2n+7}{n-5}+\frac{1-n}{n-5}\)
\(\Rightarrow A=\frac{2n+7+1-n}{n-5}\)
\(\Rightarrow A=\frac{n+8}{n-5}=\frac{n-5+13}{n-5}=1+\frac{13}{n-5}\)
Mà \(n\in Z\Rightarrow n-5\in Z\)
\(\Rightarrow n-5\in\left\{-13;-1;1;13\right\}\)
\(\Rightarrow n\in\left\{-8;4;6;18\right\}\)