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1 đó bạn
\(\left(3n+2\right)⋮\left(n-1\right)\)
\(\Rightarrow\hept{\begin{cases}\left(3n+2\right)⋮\left(n-1\right)\\\left(n-1\right)⋮\left(n-1\right)\end{cases}\Rightarrow\hept{\begin{cases}\left(3n+2\right)⋮\left(n-1\right)\\\left(3n-3\right)⋮\left(n-1\right)\end{cases}}}\)
\(\Rightarrow\left[3n+2-\left(3n-3\right)\right]⋮\left(n-1\right)\)
\(\left(3n+2-3n+3\right)⋮\left(n-1\right)\)
\(5\) \(⋮\left(n-1\right)\)
\(\Rightarrow n-1\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
Ta có bảng sau :
| \(n-1\) | 1 | -1 | 5 | -5 |
| \(n\) | 2 | 0 | 6 | -4 |
Vậy \(n\in\left\{2;0;6;-4\right\}\)
cảm ơn với tất cả các bạn