Ta có:
\(A=\frac{n+2}{n+5}=\frac{n+5-3}{n+5}=1-\frac{3}{n+5}\)
Để \(A\in Z\)thì \(\frac{3}{n+5}\in Z\)
\(\Leftrightarrow3⋮\left(n+5\right)\)
\(\Rightarrow n+5\inư\left(3\right)\)
\(\Rightarrow n+5\in\left\{1;-1;3;-3\right\}\)
Lập bảng :
n+5 | 1 | -1 | 3 | -3 |
n | -4 | -6 | -2 | -8 |
Vậy \(x\in\left\{-4;-6;-2;-8\right\}\)