Докажите равенство:
1) $\frac{1}{5 - 2\sqrt{6}} + \frac{1}{5 + 2\sqrt{6}} = 10$;
2) $\frac{2}{3\sqrt{2} + 4} - \frac{2}{3\sqrt{2} - 4} = -8$;
3) $\frac{\sqrt{2} + 1}{\sqrt{2} - 1} - \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = 4\sqrt{2}$.
$\frac{1}{5 - 2\sqrt{6}} + \frac{1}{5 + 2\sqrt{6}} = \frac{5 + 2\sqrt{6} + 5 - 2\sqrt{6}}{(5 - 2\sqrt{6})(5 + 2\sqrt{6})} = \frac{10}{5^2 - (2\sqrt{6})^2)} = \frac{10}{25 - 4 * 6} = \frac{10}{25 - 24} = \frac{10}{1} = 10$
$\frac{2}{3\sqrt{2} + 4} - \frac{2}{3\sqrt{2} - 4} = \frac{2(3\sqrt{2} - 4) - 2(3\sqrt{2} + 4)}{(3\sqrt{2} + 4)(3\sqrt{2} - 4)} = \frac{6\sqrt{2} - 8 - 6\sqrt{2} - 8}{(3\sqrt{2})^2 - 4^2} = \frac{-16}{9 * 2 - 16} = \frac{-16}{18 - 16} = \frac{-16}{2} = -8$
$\frac{\sqrt{2} + 1}{\sqrt{2} - 1} - \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = \frac{(\sqrt{2} + 1)^2 - (\sqrt{2} - 1)^2}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2 + 2\sqrt{2} + 1 - (2 - 2\sqrt{2} + 1)}{(\sqrt{2})^2 - 1^2} = \frac{3 + 2\sqrt{2} - 2 + 2\sqrt{2} - 1}{2 - 1} = \frac{4\sqrt{2}}{1} = 4\sqrt{2}$
Пожауйста, оцените решение
