ĐKXĐ: \(x\ge\frac{2}{3}\)
Đặt \(\sqrt{3x-2}=a\)
\(\Rightarrow x^3+2a^3=3xa^2\)
\(\Leftrightarrow x^3-3a^2x+2a^3=0\)
\(\Leftrightarrow\left(x-a\right)^2\left(2a+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=a\\a=-\frac{1}{2}x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3x-2}\\-\frac{1}{2}x=\sqrt{3x-2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-3x+2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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