Ta có:
\(A=\dfrac{3n+2}{n-1}=\dfrac{\left(3n-3\right)+5}{n-1}=\dfrac{3n-3}{n-1}+\dfrac{5}{n-1}=\dfrac{3\left(n-1\right)}{n-1}+\dfrac{5}{n-1}=3+\dfrac{5}{n-1}\)
Để \(A\in Z\Rightarrow\dfrac{5}{n-1}\in Z\Rightarrow5⋮n-1\) hay \(n-1\in U\left(5\right)=\left\{\pm1;\pm2\right\}\)
Lập bảng giá trị:
| \(n-1\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
| \(n\) | \(2\) | \(0\) | \(6\) | \(-4\) |
Vậy với \(n\in\left\{-4;0;2;6\right\}\) thì \(\dfrac{3n+2}{n-1}\in Z\)
Để \(A\in Z\) thì \(3n+2⋮n-1\)
\(\Rightarrow3\left(n-1\right)+5\) \(⋮n-1\)
Vì \(3\left(n-1\right)⋮n-1\)
\(\Rightarrow5⋮n-1\)
\(\Rightarrow n-1\inƯ\left(5\right)\)
mà \(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
| \(n-1\) | 1 | \(-1\) | 5 | \(-5\) |
| \(n\) | 2 | 0 | 6 | \(-4\) |
| Kết luận | nhận | nhận | nhận | nhận |
Vậy \(n\in\left\{-4;0;2;6\right\}\).
Để \(A\in Z\Rightarrow3n+2⋮n-1\)
Ta có: \(\left\{{}\begin{matrix}3n+1⋮n-1\\n-1⋮n-1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3n+1⋮n-1\\3n-3⋮n-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3n-3+4⋮n-1\\3n-3⋮n-1\end{matrix}\right.\)
\(\Rightarrow4⋮n-1\Rightarrow n-1\inƯ\left(4\right)\)
\(\Rightarrow n-1\in\left\{\pm1;\pm2;\pm3\right\}\)
\(\Rightarrow n\in\left\{-2;-1;0;2;3;4;\right\}\)
