Ta có \(2n^2-n+2=\left(2n+1\right)\left(n-1\right)+3\)
Để phép chia là phép chia hết thì \(\left(2n+1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow n\in\left\{0;-1;1;-2\right\}\)
Ta có \(\dfrac{2n^2-n+2}{2n+1}\)\(=\)\(\dfrac{2n^2+n-2n+3-1}{2n+1}\)\(=\)\(\dfrac{n\left(2n+1\right)-\left(2n+1\right)+3}{2n+1}\)
\(=\)\(\dfrac{\left(n-1\right)\left(2n+1\right)+3}{2n+1}\)\(=\)\(n\)\(-\)1\(+\dfrac{3}{2n+1}\)
\(2n^2-n+2⋮2n+1\)
\(\Leftrightarrow\)\(\dfrac{2n^2-n+2}{2n+1}\in Z\)
\(\Leftrightarrow\)\(n-1+\dfrac{3}{2n+1}\in Z\)
\(\Leftrightarrow\)\(\dfrac{3}{2n+1}\in Z\)
\(\Leftrightarrow\)\(2n+1\inƯ\left\{3\right\}\)
\(\Rightarrow\)\(2n+1=\left\{\pm1;\pm3\right\}\)
\(\cdot2n+1=1\Leftrightarrow2n=0\Leftrightarrow n=0\)
\(\cdot2n+1=3\Leftrightarrow2n=-2\Leftrightarrow n=-1\)
\(\cdot2n+1=-3\Leftrightarrow2n=-4\Leftrightarrow n=-2\)
\(\cdot2n+1=3\Leftrightarrow2n=2\Leftrightarrow n=1\)
Vậy \(n\in\left\{0;-1;-2;1\right\}\)
