Giải
Vì \(11⋮\left(x-1\right)\)
=> \(x-1\inƯ\left(11\right)\)
Ta có : Ư ( 11 ) = { 1 ; 11 } mà \(x-1\inƯ\left(11\right)\)
=> \(x-1\in\left\{1;11\right\}\)
=> \(x\in\left\{2;12\right\}\)
Vậy \(x\in\left\{2;12\right\}\)
ta có: \(11⋮x-1\Rightarrow x-1\inƯ\left(11\right)\Rightarrow x-1\in\left\{1;11\right\}\)
\(\Rightarrow x\in\left\{2;12\right\}\)
11⋮ ( x - 1 )
=> x-1 ∈Ư(11)={1;11}
ta có bảng sau
| x-1 | 1 | 11 |
| x | 2 | 12 |
vậy x=2 hoặc x=12
