Ta có:
\(5⋮x-1\)
\(\Rightarrow x-1\inƯ\left(5\right)=\left\{1;5\right\}\) ( Vì \(x\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}x-1=1\Rightarrow x=2\\x-1=5\Rightarrow x=6\end{matrix}\right.\)
Vậy x=2 hoặc x=6
5⋮ (x-1)
➩ (x-1)∈ Ư(5)={1;5}
ta có bảng sau
| x-1 | 1 | 5 |
| x | 2 | 6 |
vậy x∈{2;6}