Đáp án:
a, (x+6) chia hết (x+1) ⇒$\frac{x+6}{x+1}$ ∈ Z
⇔$\frac{x+1+5}{x+1}$ ∈ Z
⇔1+$\frac{5}{x+1}$ ∈ Z
⇒ x+1 ∈ Ư(5)={ ±1;±5}
⇒x∈{ 0;-2;4;-9}
b, $\frac{x-5}{x+2}$=$\frac{x+2-7}{x+2}$=1-$\frac{7}{x+2}$ ∈Z
⇒x+2 ∈ Ư(7) ={ ±1;±7}
⇒x ∈{ -1;-3;5;-9}
c, $\frac{2x-1}{x+2}$ =$\frac{2(x+2)-5}{x+2}$ =2-$\frac{5}{x+2}$ ∈ N
⇒x+2 ∈ Ư (5) ={ ±1;±5} và $\frac{5}{x+2}$ ≤2
x+2=1⇔x=-1 (tm)
x+2=-1⇔x=-3 (tm)
x+2=5⇔x=3 (tm)
x+2=-5⇔x=-7 (tm)