Example 29. Solve the equation

$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$
Let's denote $\sqrt{x-1}=t$, then we have $t \geqslant 0$ and $x=t^{2}+1$. By substituting the unknown in equation (17), we get

$$
\begin{aligned}
& \left\{\begin{array}{l}
t \geqslant 0 \\
\sqrt{t^{2}-4 t+4}+\sqrt{t^{2}-6 t+9}=1
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ t \geqslant 0 } \\
{ \sqrt { ( t - 2 ) ^ { 2 } } + \sqrt { ( t - 3 ) ^ { 2 } } = 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
t \geqslant 0 \\
|t-2|+|t-3|=1
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow 2 \leqslant t \leqslant 3 .
\end{aligned}
$$

Thus, equation (17) is equivalent to the inequalities

$$
2 \leqslant \sqrt{x-1} \leqslant 3 \Leftrightarrow 4 \leqslant x-1 \leqslant 9 \Leftrightarrow 5 \leqslant x \leqslant 10
$$

Therefore, the original equation (17) has infinitely many roots, specifically all numbers $x$ in the interval [5, 10].

Solving some irrational equations can be reduced to solving homogeneous equations or algebraic systems.