1. Solve the equation:

$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$
Solution. Using the equalities

$$
x+3-4 \sqrt{x-1}=(\sqrt{x-1}-2)^{2}
$$

and

$$
x+8-6 \sqrt{x-1}=(\sqrt{x-1}-3)^{2}
$$

The given equation reduces to the equation

$$
|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1
$$

Now, it is sufficient to consider the cases when $\sqrt{x-1} \leq 2.3$. The solution to the given equation is every $x \in [5,10]$. We leave the details to the reader as an exercise.