Điều kiện: $x\geq \frac{1}{2}$
$\Leftrightarrow x^3-6x^2+14x-12-(2x+1)\sqrt{2x-1}=0$
$\Leftrightarrow (x-2)^3+2(x-2)=\sqrt{(2x-1)^3}+2\sqrt{2x-1}$
Đặt $\left\{\begin{matrix} a=x-2\\ b=\sqrt{2x-1} \end{matrix}\right.(a\geq \frac{-3}{2};b\geq 0)$
$\Leftrightarrow a^3+2a=b^3+2b\Leftrightarrow (a-b)(a^2+ab+b^2+2)=0$
Do $a^2+ab+b^2+2>0\Rightarrow a=b\Rightarrow x-2=\sqrt{2x-1}\Rightarrow x=5(t/m)$
