1 Упростите выражение: a) √75 + √2(√8-√24); б) (√8-√5)². 2 Сократите дробь: a) 6+√6; б) √b+7.

1. a)

$$ \sqrt{75} + \sqrt{2}(\sqrt{8} - \sqrt{24}) = \sqrt{25 \cdot 3} + \sqrt{2}(\sqrt{4 \cdot 2} - \sqrt{4 \cdot 6}) = 5\sqrt{3} + \sqrt{2}(2\sqrt{2} - 2\sqrt{6}) = 5\sqrt{3} + 2 \cdot 2 - 2\sqrt{12} = 5\sqrt{3} + 4 - 2\sqrt{4 \cdot 3} = 5\sqrt{3} + 4 - 4\sqrt{3} = \sqrt{3} + 4 $$

б)

$$ (\sqrt{8} - \sqrt{5})^2 = (\sqrt{8})^2 - 2\sqrt{8}\sqrt{5} + (\sqrt{5})^2 = 8 - 2\sqrt{40} + 5 = 13 - 2\sqrt{4 \cdot 10} = 13 - 4\sqrt{10} $$

2. a)

$$ \frac{6+\sqrt{6}}{\sqrt{12} + \sqrt{2}} = \frac{6+\sqrt{6}}{\sqrt{4 \cdot 3} + \sqrt{2}} = \frac{6+\sqrt{6}}{2\sqrt{3} + \sqrt{2}} = \frac{\sqrt{6}(\sqrt{6}+1)}{\sqrt{2}(2\sqrt{\frac{3}{2}} + 1)} = \frac{\sqrt{3}(\sqrt{6}+1)}{2\sqrt{\frac{3}{2}} + 1} $$

б)

$$ \frac{\sqrt{b}+7}{49-b} = \frac{\sqrt{b}+7}{(7-\sqrt{b})(7+\sqrt{b})} = \frac{1}{7-\sqrt{b}} $$ Ответ: 1. a) $$\sqrt{3} + 4$$, б) $$13 - 4\sqrt{10}$$; 2. a) $$\frac{\sqrt{3}(\sqrt{6}+1)}{2\sqrt{\frac{3}{2}} + 1}$$, б) $$\frac{1}{7-\sqrt{b}}$$