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