2.14. Solve the equation

$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$
2.14. Answer: $5 \leqslant x \leqslant 10$. Note that

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

Therefore, the original equation can be written as

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

(we consider all roots to be positive). Let's consider all possible cases in turn.

1. $\sqrt{x-1}-2 \geqslant 0$ and $\sqrt{x-1}-3 \geqslant 0$, i.e., $x \geqslant 10$. In this case, the equation has a unique solution $x=10$.
2. $\sqrt{x-1}-2 \geqslant 0$ and $\sqrt{x-1}-3 \leqslant 0$, i.e., $5 \leqslant x \leqslant 10$. In this case, we obtain an identity, i.e., if $5 \leqslant x \leqslant 10$, then $x$ is a root of the given equation.
3. $\sqrt{x-1}-2 \leqslant 0$ and $\sqrt{x-1}-3 \leqslant 0$, i.e., $x \leqslant 5$. The equation has a unique solution $x=5$.

The case when $\sqrt{x-1}-2 \leqslant 0$ and $\sqrt{x-1}-3 \geqslant 0$ is obviously impossible.