\(\frac{n+5}{n+1}=\frac{n+1+4}{n+1}=\frac{n+1}{n+1}+\frac{4}{n+1}=1+\frac{4}{n+1}\)
Để \(\frac{4}{n+1}\in N\) thì \(n+1\in\text{Ư}\left(4\right)=\left\{1;2;4\right\}\)
\(n+1=1\Rightarrow n=0\)\(n+1=2\Rightarrow n=1\)\(n+1=4\Rightarrow n=3\)Vậy \(n\in\left\{0;1;3\right\}\)