1. If points $A$ and $B$ are on the same side of the $x$-axis, you can first find the symmetric point $B^{\prime}\left(x_{2}, -y_{2}\right)$ (or $A^{\prime}\left(x_{1}, -y_{1}\right)$) of point $B$ (or $A$) with respect to the $x$-axis, then calculate $\left|A B^{\prime}\right|$. The value of $\left|A B^{\prime}\right|$ is the minimum value of the distance from point $P$ to points $A$ and $B$. Therefore, the range is $\left[\left|A B^{\prime}\right|, +\infty\right)$.
Solution 1: The original expression is transformed into
$$
\begin{aligned}
y & =\sqrt{(x-1)^{2}+(0-2)^{2}} \\
& +\sqrt{(x+2)^{2}+(0-3)^{2}},
\end{aligned}
$$

which means the moving point is $P(x, 0)$, and the fixed points are $A(1,2), B(2,3)$. The point $B$'s reflection over the $x$-axis is $B^{\prime}(2,-3)$.

The line $l$ connecting $A B^{\prime}$ intersects the $x$-axis at $P\left(\frac{7}{5}, 0\right)$. When $x=\frac{7}{5}$, $y_{\text {min }}=|A B|=$ $\sqrt{26}$. 
Therefore, the range is $[\sqrt{26},+\infty)$.