\(x^2+x=0\Rightarrow x\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(x^2+x=0\)
\(\Rightarrow x.\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{0;-1\right\}.\)
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\(x^2+x=0\Leftrightarrow x^2+1x=0\\ \Leftrightarrow x.\left(x+1\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy...
x2 + x = 0
==> x.(x+1)=0
==> x = 0 hc x = -1
