Để \((x+8) \vdots (x+1) \Rightarrow \frac{x+8}{x+1} \epsilon Z\)
\(\Rightarrow \frac{x+8}{x+1} = \frac{(x+1)+7}{x+1} = 1+ \frac{7}{x+1}\epsilon Z\)
theo yêu cầu đề bài \(\Rightarrow \frac{7}{x+1}\epsilon Z\)
\(\Rightarrow (x+1) \epsilon\) Ư (7) \(\Rightarrow (x+1) \epsilon \left \{-7; -1;1;7 \right \}\)
\(\Rightarrow x \epsilon \left \{-8; -2;0;6 \right \}\)