Освободитесь от иррациональности в знаменателе дроби:
1) $\frac{\sqrt{5}}{\sqrt{5} - 2}$;
2) $\frac{8}{\sqrt{10} - \sqrt{2}}$;
3) $\frac{9}{\sqrt{x} + \sqrt{y}}$;
4) $\frac{2 - \sqrt{2}}{2 + \sqrt{2}}$.
$\frac{\sqrt{5}}{\sqrt{5} - 2} = \frac{\sqrt{5}(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)} = \frac{\sqrt{5}(\sqrt{5} + 2)}{(\sqrt{5})^2 - 2^2} = \frac{\sqrt{5}(\sqrt{5} + 2)}{5 - 4} = \frac{5 + 2\sqrt{5}}{1} = 5 + 2\sqrt{5}$
$\frac{8}{\sqrt{10} - \sqrt{2}} = \frac{8(\sqrt{10} + \sqrt{2})}{(\sqrt{10} - \sqrt{2})(\sqrt{10} + \sqrt{2})} = \frac{8(\sqrt{10} + \sqrt{2})}{(\sqrt{10})^2 - (\sqrt{2})^2} = \frac{8(\sqrt{10} + \sqrt{2})}{10 - 2} = \frac{8(\sqrt{10} + \sqrt{2})}{8} = \sqrt{10} + \sqrt{2}$
$\frac{9}{\sqrt{x} + \sqrt{y}} = \frac{9(\sqrt{x} - \sqrt{y})}{(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})} = \frac{9(\sqrt{x} - \sqrt{y})}{(\sqrt{x})^2 - (\sqrt{y})^2} = \frac{9(\sqrt{x} - \sqrt{y})}{x - y}$
$\frac{2 - \sqrt{2}}{2 + \sqrt{2}} = \frac{(2 - \sqrt{2})(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{(2 - \sqrt{2})^2}{2^2 - (\sqrt{2})^2} = \frac{2^2 - 2 * 2\sqrt{2} + (\sqrt{2})^2}{4 - 2} = \frac{4 - 4\sqrt{2} + 2}{2} = \frac{6 - 4\sqrt{2}}{2} = \frac{2(3 - 2\sqrt{2})}{2} = 3 - 2\sqrt{2}$
Пожауйста, оцените решение
