\(\dfrac{x+8}{x+1}=1+\dfrac{7}{x+1}\)
\(x+8=x+1+7\)
\(\left(x+1\right)+7⋮\left(x+1\right)\)
=> \(x+1\inƯ\left(7\right)\)
\(Ư\left(7\right)=\left\{1;-1;7;-7\right\}\)
=> \(x=\left\{0;-2;6;-8\right\}\)
\(x+8⋮x+1\)
\(\Rightarrow\left(x+8\right)-\left(x+1\right)⋮x+1\)
\(\Rightarrow x+8-x-1⋮x+1\)
\(\Rightarrow7⋮x+1\)
\(\Rightarrow x+1\in\left\{1;7;-1;-7\right\}\)
\(\Rightarrow x\in\left\{0;6;-2;-8\right\}\)