Освободить от иррациональности в знаменателе: 1) 1/(1-√2) 2) a/(√a+√b) 3) 15/(2√5+5) 4) 3/√6 Сократить дробь: 5) (m+√6)/(6-m²) 6) (√2+2)/√2 Разложить на множители: 7) √33+√22 8) 10-9a² Упростить выражение: 9) 5√24m-4√48m-√12m 10) 2√72-√50-2√8

1) $$rac{1}{1-\sqrt{2}} = \frac{1(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})} = \frac{1+\sqrt{2}}{1-2} = \frac{1+\sqrt{2}}{-1} = -1-\sqrt{2}$$

2) $$\frac{a}{\sqrt{a}+ \sqrt{b}} = \frac{a(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})} = \frac{a(\sqrt{a}-\sqrt{b})}{a-b}$$

3) $$\frac{15}{2\sqrt{5}+5} = \frac{15(2\sqrt{5}-5)}{(2\sqrt{5}+5)(2\sqrt{5}-5)} = \frac{15(2\sqrt{5}-5)}{20-25} = \frac{15(2\sqrt{5}-5)}{-5} = -3(2\sqrt{5}-5) = -6\sqrt{5}+15 = 15-6\sqrt{5}$$

4) $$\frac{3}{\sqrt{6}} = \frac{3\sqrt{6}}{\sqrt{6}\sqrt{6}} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2}$$

5) $$\frac{m+\sqrt{6}}{6-m^2} = \frac{m+\sqrt{6}}{(\sqrt{6}-m)(\sqrt{6}+m)} = \frac{1}{\sqrt{6}-m}$$

6) $$\frac{\sqrt{2}+2}{\sqrt{2}} = \frac{\sqrt{2}(\sqrt{2}+2)}{\sqrt{2}\sqrt{2}} = \frac{2+2\sqrt{2}}{2} = 1+\sqrt{2}$$

7) $$\sqrt{33}+\sqrt{22} = \sqrt{11\cdot3} + \sqrt{11\cdot2} = \sqrt{11}(\sqrt{3}+\sqrt{2})$$

8) $$10-9a^2 = (\sqrt{10})^2 - (3a)^2 = (\sqrt{10}-3a)(\sqrt{10}+3a)$$

9) $$5\sqrt{24m} - 4\sqrt{48m} - \sqrt{12m} = 5\sqrt{4\cdot6m} - 4\sqrt{16\cdot3m} - \sqrt{4\cdot3m} = 5\cdot2\sqrt{6m} - 4\cdot4\sqrt{3m} - 2\sqrt{3m} = 10\sqrt{6m} - 16\sqrt{3m} - 2\sqrt{3m} = 10\sqrt{6m} - 18\sqrt{3m}$$

10) $$2\sqrt{72} - \sqrt{50} - 2\sqrt{8} = 2\sqrt{36\cdot2} - \sqrt{25\cdot2} - 2\sqrt{4\cdot2} = 2\cdot6\sqrt{2} - 5\sqrt{2} - 2\cdot2\sqrt{2} = 12\sqrt{2} - 5\sqrt{2} - 4\sqrt{2} = 3\sqrt{2}$$