Question

Tìm x : x\(^2\)+x=0

Answer 1

\(x^2+x=0\)

\(\Rightarrow x\left(x+1\right)=0\)

\(\Rightarrow x=0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

Answer 2

\(x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Answer 3

Ta có:

\(x^2+x=0\)

\(\Rightarrow x.\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{0;-1\right\}\)

Answer 4

Ta có : x²+x =0

=> x.(x+1) =0

=>\(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy x=-1 hoặc x=0