Example 4. Solve the equation:
$$
\sqrt{x^{2}-5 x+1}+\sqrt{x^{2}-6 x+6}=2 \text{. }
$$
$$
\begin{array}{l}
\text { Solve } \sqrt{x^{2}-5 x+1}+\sqrt{x^{2}-6 x+6}=2 , \\
\text { and }\left(x^{2}-5 x+1\right)=-\left(x^{2}-6 x+6\right) \\
\quad=x-5,
\end{array}
$$
(1) We get $\sqrt{x^{2}-5 x+1}-\sqrt{x^{2}-6 x+6}=\frac{x-5}{2}$,

Then $2 \sqrt{x^{2}-5 x+1}-2 \sqrt{x^{2}-6 x+6}$
$$
\begin{array}{l}
=x-5 . \\
\quad \text { (1) } \times 2+\text { (3) gives } \\
\quad 4 \sqrt{x^{2}-5 x+1}=x-1 .
\end{array}
$$

Squaring both sides, after rearrangement we get
$$
(5 x-1)(x-5)=0 \text {. }
$$

Solving equation (4) gives $x_{1}=5, x_{2}=\frac{1}{5}$.
Substituting into (1) for verification, $x=\frac{1}{5}$ is an extraneous root and is discarded, $x=5$ is the solution to the original equation.