$2x^2-(3x-2)=x\sqrt[]{3x-2}$
đặt $\sqrt{3x-2}=t \Longrightarrow x=\dfrac{t^2+2}{3}$
$pt\Longleftrightarrow 2.\dfrac{(t^2+2)^2}{9}-t^2=\dfrac{t^2+2}{3}.t\\
\Longleftrightarrow 2(t^4+4t^2+4)-9t^2=(3t^2+6)t\\
\Longleftrightarrow 2t^4+8t^2+8-9t^2=3t^3+6t\\
\Longleftrightarrow 2t^4-3t^3-t^2-6t+8=0$
dễ rồi nhờ =))