\(\dfrac{1}{2}x^3-49x=0\)
\(\dfrac{1}{2}x^2.x-49x=0\)
\(x.\left(\dfrac{1}{2}x^2-49\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}x^2-49=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}x^2=49\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=98\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=7\sqrt{2}\end{matrix}\right.\)
Vậy \(x\in\left\{0,7\sqrt{2}\right\}\)
