Ta có : \(\dfrac{3n+1}{n+1}=\dfrac{3n+3-2}{n+1}=\dfrac{3\left(n+1\right)-2}{n+1}=3-\dfrac{2}{n+1}\)
Để \(\dfrac{3n+1}{n+1}\) đạt giá trị nguyên thì \(\dfrac{2}{n+1}\) đạt giá trị nguyên
\(\Leftrightarrow2⋮n+1\)
\(\Rightarrow n+1\inƯ_{\left(2\right)}\)
Mà \(Ư_{\left(2\right)}\in\left\{1;-1;2;-2\right\}\)
\(\Rightarrow n+1\in\left\{1;-1;2;-2\right\}\)
Ta có bảng sau :
| \(n+1\) | \(1\) | \(-1\) \(\) | \(2\) | \(-2\) |
| \(n\) | \(0\) | \(-2\) | \(1\) | \(-3\) |
Vậy \(n\in\left\{0;-2;1;-3\right\}\)
\(\dfrac{3n+1}{n+1}\)
\(\dfrac{3n+3-2}{n+1}\)
\(\dfrac{3\left(n+1\right)}{n+1}-\dfrac{2}{n+1}\)
\(3-\dfrac{2}{n+1}\)
Để \(\dfrac{3n+1}{n+1}\) đạt nguyên
\(\Leftrightarrow\dfrac{2}{n+1}\)nguyên
\(\Leftrightarrow\left(n+1\right)\in U\left(2\right)=\left\{\pm1;\pm2\right\}\)
Ta có bảng sau:
| n+1 | 1 | -1 | 2 | -2 |
| n | 0(loại) | -2 | 1 | -3 |
Vậy .....................
