$\sqrt{x^2+x+1}+\sqrt{x^2-3x+1}=2$
$⇔x^2+x+1+x^2-3x+1+2\sqrt{(x^2+x+1)(x^2-3x+1)}=4$
$⇔2x^2-2x+2+2\sqrt{x^4-3x^3+x^2+x^3-3x^2+x+x^2-3x+1}=4$
$⇔\sqrt{x^4-2x^3-x^2-2x+1}=2-x^2+x-1$
$⇔x^4-2x^3-x^2-2x+1=(-x^2+x+1)^2$
$⇔x^4-2x^3-x^2-2x+1=x^2+x^4+1+2x-2x^2-2x^3$
$⇔x^4-x^4-2x^3+2x^3-x^2-x^2+2x^2-2x-2x=0$
$⇔-4x=0$
$⇔x=0$
Vậy $x=0$