\(\left(3n+2\right)⋮\left(n-1\right)\)
\(\Rightarrow\hept{\begin{cases}\left(3n+2\right)⋮\left(n-1\right)\\3.\left(n-1\right)⋮\left(n-1\right)\end{cases}\Rightarrow\hept{\begin{cases}\left(3n+2\right)⋮\left(n-1\right)\\\left(3n-3\right)⋮\left(n-1\right)\end{cases}}}\)
\(\Rightarrow3n+2-\left(3n-3\right)⋮n-1\)
\(3n+2-3n+3⋮n-1\)
\(5⋮n-1\)
\(\Rightarrow n-1\inƯ\left(5\right)=\left\{1;5\right\}\)
Ta có bảng sau :
| \(n-1\) | 1 | 5 |
| \(n\) | 2 | 6 |
Vậy \(n\in\left\{2;6\right\}\)
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