Ta có:
\(\frac{3n^2+1}{n+2}=\frac{3n\left(n+2\right)-5}{n+2}=\frac{3n\left(n+2\right)}{n+2}-\frac{5}{n+2}=3n-\frac{5}{n+2}\)
Để phân số \(\frac{3n^2+1}{n+2}\in Z\)\(\Rightarrow3n-\frac{5}{n+2}\in Z\)
Mà \(3n\in Z\Rightarrow\left(n+2\right)\inƯ\left(5\right)\)
*\(\orbr{\begin{cases}n+2=1\\n+2=-1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}n=1-2\\n=-1-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}n=-1\\n=-3\end{cases}}\)
*\(\orbr{\begin{cases}n+2=5\\n+2=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}n=5-2\\n=-5-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}n=3\\n=-7\end{cases}}\)
Vậy \(n\in\left(-7;-3;-1;3\right)\)
