ĐKXĐ: ...
Trừ vế cho vế:
\(2x=4y-2\Rightarrow x=2y-1\)
\(\Rightarrow2y-1-2\sqrt{3y-2}=2y-3\sqrt[3]{4y-3}\)
\(\Leftrightarrow2\sqrt{3y-2}+1-3\sqrt[3]{4y-3}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{3y-2}=a\ge0\\\sqrt[3]{4y-3}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a+1-3b=0\\4a^2-3b^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{3b-1}{2}\\4a^2-3b^3=1\end{matrix}\right.\)
\(\Rightarrow\left(3b-1\right)^2-3b^3=1\)
\(\Leftrightarrow3b^2-2b-b^3=0\)
\(\Leftrightarrow b\left(b^2-3b+2\right)=0\Rightarrow\left[{}\begin{matrix}b=0\\b=1\\b=2\end{matrix}\right.\) \(\Rightarrow y=...\Rightarrow x=...\)
