Có $\frac{x-6}{x+1}$ =$\frac{x+1-7}{x+1}$ =1-$\frac{7}{x+1}$ ∈Z
⇔$\frac{7}{x+1}$ ∈ Z
⇔$\left \{ {{x∈Z,x khác -1} \atop {x+1∈Ư(7)}} \right.$
⇔$\left \{ {{x∈Z,x khác -1} \atop {x+1∈{-7;-1;1;7}}} \right.$
⇔$\left \{ {{x∈Z,x khác -1} \atop {x∈{-8;-2;0;6}}} \right.$
Vậy x∈{-8;-2;0;6} thì x-6 chia hết x-1