Let's simplify the following expression:

$$
\frac{(x-1) \sqrt{x^{2}-6 x+8}}{(x-2) \sqrt{x^{2}-4 x+3}}+\frac{(x-5) \sqrt{x^{2}-4 x+3}}{(x-3) \sqrt{x^{2}-7 x+10}}
$$
Each expression under the square root can be written as the product of two linear expressions. If we also bring the factors outside the root under the root, then we can simplify each term by 2 linear factors. Thus,

$$
\begin{gathered}
K=\sqrt{\frac{(x-1)^{2}(x-2)(x-4)}{(x-2)^{2}(x-1)(x-3)}}+\sqrt{\frac{(x-5)^{2}(x-1)(x-3)}{(x-3)^{2}(x-2)(x-5)}}= \\
=\frac{\sqrt{(x-1)(x-4)}+\sqrt{(x-1)(x-5)}}{\sqrt{(x-2)(x-3)}}
\end{gathered}
$$

indeed, after simplification, the two denominators turned out to be equal. By factoring, this can also be written as:

$$
K=\frac{\sqrt{x-1}(\sqrt{x-4}+\sqrt{x-5})}{\sqrt{x^{2}-5 x+6}}
$$

Anna Illyés (Budapest, Berzsenyi D. Secondary School II. grade)

Note. Several people rationalized the denominator as the last step. If the value of $K$ also had to be calculated for certain values of $x$, and these $x$ values happened to be "small" integers, then (but only then) this form would indeed be more convenient. However, the structure of the expression is more comprehensible from the above forms. We should only rationalize the denominator if we see its purpose and advantage.