Ta có : $x^3+x=0$
$=>x(x^2+1)=0$
\(=>\left[{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=0\\x^2=-1\left(vo-ly\right)\end{matrix}\right.\)
$=>x=0$
\(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=0\)
0
\(x^3+x=0\Rightarrow x\left(x^2+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=-1\end{matrix}\right.\)\(x^2=-1\) (loại)
Vậy x = 0
\(x^3+x=0\)
\(\Rightarrow x\left(x^2+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=-1\left(VL\right)\end{matrix}\right.\)
Vì \(x^2\ge0\) nên \(x^2=-1\) (loại)
Vậy, \(x=0\)
