a,
b,
3/
\(\frac{2n^2-n+2}{2n+1}=\frac{2n^2+n-2n-1+3}{2n+1}=\frac{n\left(2n+1\right)-\left(2n+1\right)+3}{2n+1}=n-1+\frac{3}{2n+1}\)
Để \(2n^2-n+2⋮2n+1\Leftrightarrow2n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng:
2n+1 | 1 | -1 | 3 | -3 |
n | 0 | -1 | 1 | -2 |
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