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

Question

Вопрос:

Ответ:

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

Accepted Answer

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