\(\Leftrightarrow\sqrt[3]{x-5}+\sqrt[3]{2x-1}=\sqrt[3]{3x+2}-2\)
\(\Leftrightarrow x-5+2x-1+3\sqrt[3]{\left(x-5\right)\left(2x-1\right)}\left(\sqrt[3]{x-5}+\sqrt[3]{2x-1}\right)=3x+2-8-6\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)\)
\(\Leftrightarrow3\sqrt[3]{\left(x-5\right)\left(2x-1\right)}\left(\sqrt[3]{x-5}+\sqrt[3]{2x-1}\right)+6\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)=0\)
\(\Leftrightarrow3\sqrt[3]{\left(x-5\right)\left(2x-1\right)}\left(\sqrt[3]{3x+2}-2\right)+6\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt[3]{3x+2}-2\right)\left(\sqrt[3]{\left(x-5\right)\left(2x-1\right)}+2\sqrt[3]{3x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=8\\\left(x-5\right)\left(2x-1\right)=-8\left(3x+2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\2x^2+13x+21=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy ...