第63問 答え

$$\begin{align}&(1+\sqrt{6})^2 - \frac{\sqrt{8}+10\sqrt{3}}{\sqrt{2}}\\&=1 + 2\sqrt{6} + 6 - \frac{\sqrt{2} \times (2\sqrt{2} + 10 \sqrt{3})}{\sqrt{2} \times \sqrt{2}}\\&=7 + 2\sqrt{6} - \frac{4 + 10 \sqrt{6}}{2}\\&=7 + 2\sqrt{6} - 2 - 5 \sqrt{6}\\&=5 - 3\sqrt{6}\end{align}$$