Освободитесь от иррациональности в знаменателе дроби: a) $$rac{x}{x + \sqrt{y}}$$; б) $$rac{b}{a - \sqrt{b}}$$; в) $$rac{4}{\sqrt{10} - \sqrt{2}}$$; г) $$rac{12}{\sqrt{3} + \sqrt{6}}$$; д) $$rac{9}{3 - 2\sqrt{2}}$$; e) $$rac{14}{1 + 5\sqrt{2}}$$.

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ФИО:Телефон:Емаил:Полное описание сути нарушения прав (почему распространение данной информации запрещено Правообладателем):

Accepted Answer

a) $$rac{x}{x + \sqrt{y}} = \frac{x(x - \sqrt{y})}{(x + \sqrt{y})(x - \sqrt{y})} = \frac{x(x - \sqrt{y})}{x^2 - y}$$
б) $$rac{b}{a - \sqrt{b}} = \frac{b(a + \sqrt{b})}{(a - \sqrt{b})(a + \sqrt{b})} = \frac{b(a + \sqrt{b})}{a^2 - b}$$
в) $$rac{4}{\sqrt{10} - \sqrt{2}} = \frac{4(\sqrt{10} + \sqrt{2})}{(\sqrt{10} - \sqrt{2})(\sqrt{10} + \sqrt{2})} = \frac{4(\sqrt{10} + \sqrt{2})}{10 - 2} = \frac{4(\sqrt{10} + \sqrt{2})}{8} = \frac{\sqrt{10} + \sqrt{2}}{2}$$
г) $$rac{12}{\sqrt{3} + \sqrt{6}} = \frac{12(\sqrt{3} - \sqrt{6})}{(\sqrt{3} + \sqrt{6})(\sqrt{3} - \sqrt{6})} = \frac{12(\sqrt{3} - \sqrt{6})}{3 - 6} = \frac{12(\sqrt{3} - \sqrt{6})}{-3} = -4(\sqrt{3} - \sqrt{6}) = 4(\sqrt{6} - \sqrt{3})$$
д) $$rac{9}{3 - 2\sqrt{2}} = \frac{9(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} = \frac{9(3 + 2\sqrt{2})}{9 - 8} = \frac{9(3 + 2\sqrt{2})}{1} = 9(3 + 2\sqrt{2}) = 27 + 18\sqrt{2}$$
е) $$rac{14}{1 + 5\sqrt{2}} = \frac{14(1 - 5\sqrt{2})}{(1 + 5\sqrt{2})(1 - 5\sqrt{2})} = \frac{14(1 - 5\sqrt{2})}{1 - 50} = \frac{14(1 - 5\sqrt{2})}{-49} = \frac{2(1 - 5\sqrt{2})}{-7} = \frac{2(5\sqrt{2} - 1)}{7}$$