Let's calculate the value of the following expression:

$$
\frac{\sqrt{6}+2 \sqrt{2}}{3-\sqrt{6}-\sqrt{3}+\sqrt{2}}
$$
It is visible that the denominator can be transformed into a product:

$$
(\sqrt{9}-\sqrt{6})-(\sqrt{3}-\sqrt{2})=(\sqrt{3}-1)(\sqrt{3}-\sqrt{2})
$$

Therefore, by extending with $K-\mathrm{t}(\sqrt{3}+1)(\sqrt{3}+\sqrt{2})$, the new denominator will be $(3-1)(3-2)=2$. Performing the multiplications,

$$
\begin{aligned}
K & =\frac{1}{2}(\sqrt{6}+2 \sqrt{2})(\sqrt{3}+1)(\sqrt{3}+\sqrt{2})=\frac{1}{2}(5 \sqrt{2}+3 \sqrt{6})(\sqrt{3}+\sqrt{2})= \\
& =\frac{1}{2}(10+9 \sqrt{2}+6 \sqrt{3}+5 \sqrt{6})=\frac{1}{2}(10+\sqrt{162}+\sqrt{108}+\sqrt{150}
\end{aligned}
$$

Rounding to three decimal places already, $\sqrt{162} \approx 12.728, \sqrt{108} \approx 10.392, \sqrt{150} \approx 12.247$, so $K \approx 45.367 / 2 \approx 22.684$ (with rounding up).

The error of each of the three roots is less than $5 / 10^{4}$, so the error of the dividend is less than $15 / 10^{4}$, and finally, the error of $K$ is less than $7.5 / 10^{4}$, so rounding to three decimal places, the value of $K$ is either 22.683 or 22.684.

Ákos Szentirmai (Sopron, Széchenyi I. g. II. o. t.)

Note. If the roots in the form $K=5+\sqrt{40.5}+\sqrt{27}+\sqrt{37.5}$ are calculated from their values rounded to three decimal places, the error limit of $K$ would be $15 / 10^{4}$, because the halving of the error limit is omitted by performing the division in advance.