3-4. Solve the equation:

\[
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
\]
Solution 4. Answer: \(5 \leqslant x \leqslant 10\). Notice 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 one by one.
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 where \(\sqrt{x-1}-2 \leqslant 0\) and \(\sqrt{x-1}-3 \geqslant 0\) is obviously impossible.