\(\frac{2n^2+9n+7}{2n+1}=\frac{\left(2n^2+9n+4\right)+3}{2n+1}=\frac{\left(2n^2+n+8n+4\right)+3}{2n+1}\)
\(=\frac{n\left(2n+1\right)+4\left(2n+1\right)+3}{2n+1}=\frac{\left(n+4\right)\left(2n+1\right)+3}{2n+1}=n+4+\frac{3}{2n+1}\)
Để phân thức trên là 1 số nguyên <=> \(3⋮2n+1\Rightarrow2n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow n\in\left\{-2;-1;0;1\right\}\)