\(P=\frac{2n^2-n+2}{2n+1}=\frac{n\left(2n+1\right)-\left(2n-2\right)}{2n+1}=n-\frac{2n-2}{2n+1}\)
\(=n-\frac{2n+1-3}{2n+1}=n-1+\frac{3}{2n+1}\)
Để P nguyên thì \(\frac{3}{2n+1}\)nguyên
\(\Leftrightarrow3⋮\left(2n+1\right)\Leftrightarrow2n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Lập bảng:
| \(2n+1\) | \(1\) | \(-1\) | \(3\) | \(-3\) |
| \(n\) | \(0\) | \(-1\) | \(1\) | \(-2\) |
Vậy \(n\in\left\{-2;-1;0;1\right\}\)
#)Giải :
\(P=\frac{2n^2-n+2}{2n+1}=\frac{2n^2+n-2n-1+3}{2n+1}=\frac{n\left(2n+1\right)-\left(2n+1\right)+3}{2n+1}\)
\(=\frac{\left(2n+1\right)\left(n-1\right)+3}{2n+1}=n-1+\frac{3}{2n+1}\)
\(\Rightarrow2n+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow\orbr{\begin{cases}2n+1=-3\\2n+1=1\end{cases}\Rightarrow\orbr{\begin{cases}n=-2\\n=-1\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}2n+1=1\\2n+1=3\end{cases}\Rightarrow\orbr{\begin{cases}n=0\\n=1\end{cases}}}\)
Vậy \(n\in\left\{-2;-1;0;1\right\}\)
