đặt $\sqrt{2x^2-4x+1}=t,t \geq 0$
$pt <=> (3x-2)t=2t^2+(x^2-\frac{1}{2}-\frac{3}{2})$
$\Delta =(x-4)^2$
$=> \left[t=\frac{3x-2+(x-4)}{4}; t=\frac{3x-2-(x-4)}{4}]\right.$
thay t zô ;
kq : $x=(\frac{1}{2}+\sqrt{\frac{3}{2}};\frac{9 \pm 2\sqrt{15}}{7})$