PT đã cho tương đương với cái nè$3(x^2-2x)\sqrt{2x-1}=2(x^3+5x)-4(2x^2+1)$
$\Leftrightarrow 3x(x-2)\sqrt{2x-1}=2(x^3+5x-4x^2-2)$
$\Leftrightarrow 3x(x-2)\sqrt{2x-1}=2(x-1)^2(x-2)$
$\Leftrightarrow (x-2)[3x\sqrt{2x-1}-2(x-1)^2]=0$
TH1: $x=2$ thỏa mãn
TH2: $3x\sqrt{2x-1}=2(x-1)^2$
Đặt $\sqrt{2x-1}=y\Rightarrow (x-1)^2=x^2-y^2$
$\Rightarrow 2x^2-2y^2-3xy=0$
$\Rightarrow (2x+y)(x-2y)=0$
Do $x>0 , y\ge0\Rightarrow 2x+y>0\Rightarrow x=2y$
$\Rightarrow x=2\sqrt{2x-1}$
$\Rightarrow x^2=8x-4$
$\Rightarrow (x-4)^2=12$
$\Rightarrow x=4\pm2\sqrt3$