56. $$\frac{\sqrt{3}}{\sqrt{6}+2}-\frac{\sqrt{2}}{4-\sqrt{3}}$$

56. Решим пример:

• $$\frac{\sqrt{3}}{\sqrt{6}+2}-\frac{\sqrt{2}}{4-\sqrt{3}} = \frac{\sqrt{3}(\sqrt{6}-2)}{(\sqrt{6}+2)(\sqrt{6}-2)}-\frac{\sqrt{2}(4+\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})} = \frac{\sqrt{3}(\sqrt{6}-2)}{6-4}-\frac{\sqrt{2}(4+\sqrt{3})}{16-3} = \frac{\sqrt{3}(\sqrt{6}-2)}{2}-\frac{\sqrt{2}(4+\sqrt{3})}{13} = \frac{3\sqrt{2}-2\sqrt{3}}{2} - \frac{4\sqrt{2}+\sqrt{6}}{13} = \frac{13(3\sqrt{2}-2\sqrt{3}) - 2(4\sqrt{2}+\sqrt{6})}{26} = \frac{39\sqrt{2} - 26\sqrt{3} - 8\sqrt{2} - 2\sqrt{6}}{26} = \frac{31\sqrt{2} - 26\sqrt{3} - 2\sqrt{6}}{26}$$

Ответ:$$\frac{31\sqrt{2} - 26\sqrt{3} - 2\sqrt{6}}{26}$$