Solve the equation

$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$
Solution

We have

$$
\begin{aligned}
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}} & =\sqrt{(\sqrt{x-1}-2)^{2}}+\sqrt{(\sqrt{x-1}-3)^{2}} \\
& =|\sqrt{x-1}-2|+|\sqrt{x-1}-3|
\end{aligned}
$$

Since the distance between 2 and 3 is 1, we can only have $|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$ when $\sqrt{x-1}$ lies between these two numbers, that is,

$$
2 \leq \sqrt{x-1} \leq 3
$$

This is equivalent to $4 \leq x-1 \leq 9$. Therefore, the solution set is $\{x \in \mathbb{R} \mid 5 \leq x \leq 10\}$.