x + 12 \(⋮\)x - 1
=> x-1 \(⋮\)x-1
=> ( x+12) -( x-1) \(⋮\)x-1
=> x+ 12 -x + 1 \(⋮\)x-1
=>13 \(⋮\)x-1
=> x-1 \(\in\)Ư(13)={ 1;13}
=> x \(\in\){2; 14}
Vậy....
x + 12 ⋮ x - 1
=> x-1 ⋮ x-1
=> ( x+12) -( x-1) ⋮ x-1
=> x+ 12 -x + 1 ⋮ x-1
=>13 ⋮ x-1
=> x-1 ∈ Ư(13)={ 1;13}
=> x ∈ {2; 14}
Vậy x ∈ {2; 14}