ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt[6]{3-4x}=a\ge0\\2x=b\end{matrix}\right.\)
\(\Rightarrow a^3+2a^2b-3b^3=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+3ab+3b^2\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt[6]{3-4x}=2x\)
\(\Leftrightarrow2x-1+1-\sqrt[6]{3-4x}=0\)
\(\Leftrightarrow2x-1+\frac{2\left(2x-1\right)}{\left(1+\sqrt[6]{3-4x}+\sqrt[3]{3-4x}\right)\left(\sqrt{3-4x}+1\right)}=0\)
\(\Leftrightarrow\left(2x-1\right)\left(1+\frac{2}{\left(1+\sqrt[6]{3-4x}+\sqrt[3]{3-4x}\right)\left(\sqrt{3-4x}+1\right)}\right)=0\)
\(\Leftrightarrow x=\frac{1}{2}\)