Tìm n ∈ \(\mathbb{Z}\) để \(\dfrac{2n+1}{n-2}\in\) \(\mathbb{Z}\).
\(n\in\left\{-3;1;3;7\right\}\)\(n\in\left\{1;3\right\}\)\(n\in\left\{-1;1;3;5\right\}\)\(n\in\left\{1;3;7\right\}\)Hướng dẫn giải:Để \(\dfrac{2n+1}{n-2}\in\)\(\mathbb{Z}\) thì \(\dfrac{2n+1}{n-2}=\dfrac{2n-4+5}{n-2}=2+\dfrac{5}{n-2}\in\) \(\mathbb{Z}\).
⇒ \(\dfrac{5}{n-2}\in\) \(\mathbb{Z}\) ⇒ \(n-2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
| n - 2 | -5 | -1 | 1 | 5 |
| n | -3 | 1 | 3 | 7 |
