Để \(A=\frac{5n+1}{n+1}\in Z\) \(\Leftrightarrow5n+1⋮n+1\)
\(\Leftrightarrow\) \(5n+1-5\left(n+1\right)⋮n+1\) (Vì 5(n+1)⋮n+1)
\(\Leftrightarrow5n+1-5n-5⋮n+1\)
\(\Leftrightarrow-4⋮n+1\)
\(\Rightarrow n+1\in\) Ư\(\left(-4\right)=\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow n\in\left\{0;1;3;-2;-3;-5\right\}\)
Mà \(n\in N\) nên \(n\in\left\{0;1;3\right\}\)
Vậy để \(A\) nguyên thì \(n\in\left\{0;1;3\right\}\) (\(n\in N\))