ĐKXĐ: \(x\ge\frac{2}{3}\)
Đặt \(\left\{{}\begin{matrix}\sqrt[4]{3x-2}=a\\\sqrt[4]{x+2}=b\end{matrix}\right.\)
\(\Rightarrow2a^2+b^2=3ab\Leftrightarrow2a^2-3ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[4]{3x-2}=\sqrt[4]{x+2}\\2\sqrt[4]{3x-2}=\sqrt[4]{x+2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=x+2\\16\left(3x-2\right)=x+2\end{matrix}\right.\)
