648. a) \(\frac{\sqrt{3}+2\sqrt{2}}{\sqrt{7}+\sqrt{24}}+\frac{\sqrt{3}-2\sqrt{2}}{\sqrt{7}-\sqrt{24}}\)

Решение:

Приведем дроби к общему знаменателю:

\( \frac{\sqrt{3}+2\sqrt{2}}{\sqrt{7}+\sqrt{24}} = \frac{(\sqrt{3}+2\sqrt{2})(\sqrt{7}-\sqrt{24})}{(\sqrt{7}+\sqrt{24})(\sqrt{7}-\sqrt{24})} = \frac{\sqrt{21}-\sqrt{72}+2\sqrt{14}-2\sqrt{48}}{7-24} \)

\( \frac{\sqrt{3}-2\sqrt{2}}{\sqrt{7}-\sqrt{24}} = \frac{(\sqrt{3}-2\sqrt{2})(\sqrt{7}+\sqrt{24})}{(\sqrt{7}-\sqrt{24})(\sqrt{7}+\sqrt{24})} = \frac{\sqrt{21}+\sqrt{72}-2\sqrt{14}-2\sqrt{48}}{7-24} \)

Сложим полученные дроби:

\( \frac{\sqrt{21}-\sqrt{72}+2\sqrt{14}-2\sqrt{48} + \sqrt{21}+\sqrt{72}-2\sqrt{14}-2\sqrt{48}}{-17} = \frac{2\sqrt{21}-4\sqrt{48}}{-17} \)

Упростим \( \sqrt{48} \): \( \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \).

\( \frac{2\sqrt{21}-4(4\sqrt{3})}{-17} = \frac{2\sqrt{21}-16\sqrt{3}}{-17} = \frac{16\sqrt{3}-2\sqrt{21}}{17} \)

Ответ: \( \frac{16\sqrt{3}-2\sqrt{21}}{17} \).