53. Сократите дробь: a) √10-√2 5-√5 б) √6-√2 √6+√2

a) \$$\frac{\sqrt{10}-\sqrt{2}}{5-\sqrt{5}}$$\

• Преобразуем числитель и знаменатель: $$\frac{\sqrt{10}-\sqrt{2}}{5-\sqrt{5}} = \frac{\sqrt{2}(\sqrt{5}-1)}{\sqrt{5}(\sqrt{5}-1)}$$
• Сократим дробь: $$\frac{\sqrt{2}(\sqrt{5}-1)}{\sqrt{5}(\sqrt{5}-1)} = \frac{\sqrt{2}}{\sqrt{5}} = \sqrt{\frac{2}{5}} = \frac{\sqrt{10}}{5}$$

Ответ: $$\frac{\sqrt{10}}{5}$$

б) \$$\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$$\

• Умножим числитель и знаменатель на сопряженное выражение \$$\sqrt{6}-\sqrt{2}$$\$$: $$\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} = \frac{(\sqrt{6}-\sqrt{2})(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})} = \frac{(\sqrt{6}-\sqrt{2})^2}{(\sqrt{6})^2-(\sqrt{2})^2} = \frac{(\sqrt{6})^2 - 2\sqrt{6}\sqrt{2} + (\sqrt{2})^2}{6-2} = \frac{6 - 2\sqrt{12} + 2}{4} = \frac{8 - 2\sqrt{4\times3}}{4} = \frac{8 - 4\sqrt{3}}{4} = 2-\sqrt{3}$$

Ответ: $$2-\sqrt{3}$$