Đáp án:
Giải thích các bước giải:
Để $\frac{n^{2}-2n-22}{n+3}$ $∈$ $Z$
$⇒n^{2}-2n-22$ $\vdots$ $n+3$
$⇒n^{2}+3n-5n-22$ $\vdots$ $n+3$
$⇒n.(n+3)-5n-22$ $\vdots$ $n+3$
$⇒-5n-22$ $\vdots$ $n+3$
$⇒-5n-15+15-22$ $\vdots$ $n+3$
$⇒-5.(n+3)-7$ $\vdots$ $n+3$
$⇒7$ $\vdots$ $n+3$
$⇒n+3=${$7;1;-1;-7$}
$⇒n=${$4;-2;-4;-10$}