2. Đặt $\sqrt{x^{2}-x+3}=y, y>0$pt TT $\begin{cases}x^{2}-xy+y-3=0 \\ y^{2}-x^{2}+x-3=0 \end{cases}$
$\Rightarrow (x-y)(2x+y-1)=0 \Leftrightarrow x=y$ or $y=1-2x$
+) $x=y \Rightarrow x=3 (t/m)$
+) $y=1-2x$
$\Leftrightarrow x^{2}-x+3=1-4x+4x^{2} (x\leq \frac{1}{2})$
$\Leftrightarrow x=\frac{3\pm \sqrt{33}}{6}$ k/h vs đk $\Rightarrow x=\frac{3-\sqrt{33}}{6}$