12. (2001 National High School Competition Question) The range of the function $y=x+\sqrt{x^{2}-3 x+2}$ is
12. Fill $\left[1, \frac{3}{2}\right) \cup[2,+\infty)$. Reason: From $y=x+\sqrt{x-3 x+2}$, we have $\sqrt{x^{2}-3 x+2}=y-x \geqslant 0$. Squaring both sides of this equation gives $(2 y-3) x=y^{2}-2$. Thus, $y \neq \frac{3}{2}$ and $x=\frac{y^{2}-2}{2 y-3}$. From $y-x=y-\frac{y^{2}-2}{2 y-3} \geqslant 0$, we get $\frac{y^{2}-3 y+2}{2 y-3} \geqslant 0$, which implies $1 \leqslant y<\frac{3}{2}$ or $y \geqslant 2$.

For any $y \geqslant 2$, let $x=\frac{y^{2}-2}{2 y-3}$. It is easy to see that $x \geqslant 2$, so $x^{2}-3 x+2 \geqslant 0$ and $y=x+\sqrt{x^{2}-3 x+2}$. For any $1 \leqslant y<\frac{3}{2}$, similarly let $x=\frac{y^{2}-2}{2 y-3}$. It is easy to see that $x \leqslant 1$. Thus, $x^{2}-3 x+2 \geqslant 0$ and $y=x+\sqrt{x^{2}-3 x+2}$. Therefore, the range of the function is $\left[1, \frac{3}{2}\right) \cup[2,+\infty)$.