11. The range of the function $y=x+\sqrt{x^{2}-3 x+2}$ is
$$
\left[1, \frac{3}{2}\right) \cup[2,+\infty)
$$
11. 【Analysis and Solution】First, square to remove the square root.

From the given, we have $(y-x)^{2}=x^{2}-3 x+2$, then $x=\frac{y^{2}-2}{2 y-3}$.
By $y \geqslant x$, we get $y \geqslant \frac{y^{2}-2}{2 y-3}$.
Solving this, we get $1 \leqslant y<\frac{3}{2}$, or $y \geqslant 2$.
Since $\sqrt{x^{2}-3 x+2}$ can reach the lower bound 0, the range of the function is $\left[1, \frac{3}{2}\right) \cup[2,+\infty)$.