ĐKXĐ: \(x\le\frac{5}{2}\)
\(\Leftrightarrow2\left(4x^2+1\right)x=\left(6-2x\right)\sqrt{5-2x}\)
\(\Leftrightarrow8x^3+2x=\left(5-2x\right)\sqrt{5-2x}+\sqrt{5-2x}\)
\(\Leftrightarrow\left(2x\right)^3+2x=\left(\sqrt{5-2x}\right)^3+\sqrt{5-2x}\)
Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt{5-2x}=b\end{matrix}\right.\)
\(\Leftrightarrow a^3+a=b^3+b\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\) (do \(a^2+ab+b^2+1=\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}+1>0\))
\(\Leftrightarrow2x=\sqrt{5-2x}\) (\(x\ge0\))
\(\Leftrightarrow4x^2+2x-5=0\) \(\Rightarrow x=\frac{-1+\sqrt{21}}{4}\)
