Để M xác định
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{6-3x}\ge0\\\sqrt[3]{x^2-3x}\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6-3x\ge0\\x\left(x-3\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x\ne0\\x\ne3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x\ne0\end{matrix}\right.\)
Vậy.....