$a$) Để $M$ $∈$ $Z$ thì $x+6 \vdots x+1$
$⇔ x+6 - (x+1) \vdots x+1$
$⇔ x+6 - x - 1 \vdots x+1$
$⇔ 5 \vdots x+1$
$⇒$ $x+1$ $∈$ Ư($5$)={$±1;±5$}
$⇔$ $x$ $∈$ {$-6;-2;0;4$}
Vậy $x$ $∈$ {$-6;-2;0;4$}
$b$) Để $N$ $∈$ $Z$ thì $2x+6 \vdots x+2$
$⇔ 2x+6 - 2(x+2) \vdots x+2$
$⇔ 2x+6 - 2x - 4 \vdots x+2$
$⇔ 2 \vdots x+2$
$⇒$ $x+2$ $∈$ Ư($2$)={$±1;±2$}
$⇔$ $x$ $∈$ {$-4;-3;-1;0$}
Vậy $x$ $∈$ {$-4;-3;-1;0$}
$c$) Để $N$ $∈$ $Z$ thì $x+2 \vdots x-5$
$⇔ x+2 - (x-5) \vdots x-5$
$⇔ x+2 - x + 5\vdots x-5$
$⇔ 7 \vdots x-5$
$⇒$ $x-5$ $∈$ Ư($2$)={$±1;±7$}
$⇔$ $x$ $∈$ {$-2;4;6;12$}
Vậy $x$ $∈$ {$-2;4;6;12$}
