Освободитесь от иррациональности в знаменателе дроби:
а) $\frac{x}{x + \sqrt{y}}$;
б) $\frac{b}{a - \sqrt{b}}$;
в) $\frac{4}{\sqrt{10} - \sqrt{2}}$;
г) $\frac{12}{\sqrt{3} + \sqrt{6}}$;
д) $\frac{9}{3 - 2\sqrt{2}}$;
е) $\frac{14}{1 + 5\sqrt{2}}$.
$\frac{x}{x + \sqrt{y}} = \frac{x(x - \sqrt{y})}{(x - \sqrt{y})(x + \sqrt{y})} = \frac{x(x - \sqrt{y})}{x^2 - y}$
$\frac{b}{a - \sqrt{b}} = \frac{b(a + \sqrt{b})}{(a - \sqrt{b})(a + \sqrt{b})} = \frac{b(a + \sqrt{b})}{a^2 - b}$
$\frac{4}{\sqrt{10} - \sqrt{2}} = \frac{4(\sqrt{10} + \sqrt{2})}{(\sqrt{10} - \sqrt{2})(\sqrt{10} + \sqrt{2})} = \frac{4(\sqrt{10} + \sqrt{2})}{10 - 2} = \frac{4(\sqrt{10} + \sqrt{2})}{8} = \frac{\sqrt{10} + \sqrt{2}}{2}$
$\frac{12}{\sqrt{3} + \sqrt{6}} = \frac{12(\sqrt{3} - \sqrt{6})}{(\sqrt{3} - \sqrt{6})(\sqrt{3} + \sqrt{6})} = \frac{12(\sqrt{3} - \sqrt{6})}{3 - 6} = \frac{12(\sqrt{3} - \sqrt{6})}{-3} = -4(\sqrt{3} - \sqrt{6}) = 4(\sqrt{6} - \sqrt{3})$
$\frac{9}{3 - 2\sqrt{2}} = \frac{9(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} = \frac{9(3 + 2\sqrt{2})}{9 - 4 * 2} = \frac{9(3 + 2\sqrt{2})}{9 - 8} = 9(3 + 2\sqrt{2})$
$\frac{14}{1 + 5\sqrt{2}} = \frac{14(1 - 5\sqrt{2})}{(1 - 5\sqrt{2})(1 + 5\sqrt{2})} = \frac{14(1 - 5\sqrt{2})}{1 - 25 * 2} = \frac{14(1 - 5\sqrt{2})}{1 - 50} = -\frac{14(1 - 5\sqrt{2})}{49} = \frac{2(5\sqrt{2} - 1)}{7}$
Пожауйста, оцените решение
