Ta có : \(2n^2+5n-1\)
\(=2n^2-n+6n-3+2\)
\(=n\left(2n-1\right)+3\left(2n-1\right)+2\)
\(=\left(n+3\right)\left(2n-1\right)+2\)
Để \(2n^2+5n-1⋮2n-1\)
\(\Leftrightarrow\left(n+3\right)\left(2n-1\right)+2⋮2n-1\)
\(\Leftrightarrow2⋮2n-1\)
Do \(n\in Z\Rightarrow2n-1\in Z\)
\(\Rightarrow2n-1\in\left\{1;-1;2;-2\right\}\)
Ta có bảng sau :
| \(2n-1\) | \(1\) | \(-1\) | \(2\) | \(-2\) |
| \(2n\) | \(2\) | \(0\) | \(3\) | \(-1\) |
| \(n\) | \(1\) | \(0\) | \(\dfrac{3}{2}\left(L\right)\) | \(-\dfrac{1}{2}\left(L\right)\) |
Vậy \(n\in\left\{0;1\right\}\Leftrightarrow2n^2+5n-1⋮2n-1\)
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