\(x+5⋮x-1\)
=>\(x-1+6⋮x-1\)
=>\(6⋮x-1\)
=>\(x-1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)
Ta có : x + 5 ⋮ x - 1
=> (x - 1) + 6 ⋮ x - 1
Vì x - 1 ⋮ x - 1 nên 6 ⋮ x - 1 => x - 1 ∈ Ư(6) ∈ { -6;-3;-2;-1;1;2;3;6}
=> x = -5;-2;-1;0;2;3;4;7
(x+5) chia hết cho (x-1)
x+1+5 chia hết cho x-1