Ta có: x2+x+1 là bội của x+1
\(\Rightarrow x^2+x+1⋮x+1\)
\(\Leftrightarrow x.x+x+1⋮x+1\)
\(\Leftrightarrow x\left(x+1\right)+1⋮x+1\)
Vì: \(\left\{{}\begin{matrix}x\left(x+1\right)+1⋮x+1\\x\left(x+1\right)⋮x+1\end{matrix}\right.\)
nên: 1 \(⋮\) x+1
\(\Rightarrow x+1\inƯ\left(1\right)=\left\{1;-1\right\}\)
\(\Rightarrow x\in\left\{0;-2\right\}\)
Vậy:................