ĐK: x\(\ge\)0
Ta có:
\(\left|x\left(x^2-3\right)\right|=x\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-3\right)=x\\x\left(x^2-3\right)=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^3-3x=x\\x^3-3x=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^3-4x=0\\x^3-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-4\right)=0\\x\left(x^2-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=-2\\x=2\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=-\sqrt{2}\\x=\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
Loại các nghiệm x=-2;x=-\(\sqrt{2}\) (vì x<0)
Vậy phương trình có 3 nghiệm x=0;x=2;x=\(\sqrt{2}\)
