Câu trả lời:
\(x^2+2x+1=0\)
\(x.x+2.x+1=0\)
\(x\left(x+2\right)+1=0\)
\(\hept{\begin{cases}x=0\\\left(x+2\right)+1=0\end{cases}}\)
\(\hept{\begin{cases}x=0\\x+2=-1\end{cases}}\)
\(\hept{\begin{cases}x=0\\x=-3\end{cases}}\)
Vậy :\(x=\hept{\begin{cases}0\\-3\end{cases}}\)