Упр.2.136 ГДЗ Дорофеев Суворова 8 класс (Алгебра)

а) x=v2,y=v8;
б) x=2-v3,y=2+v3;
в) x=v6-v3,y=v6+v3;
г) x=v5+v2,y=v5-v2.

Вычислим значения выражений $$\frac{xy}{x+y}$$ и $$\frac{x-y}{xy}$$ для каждого случая.

1. а) При $$x=\sqrt{2},\ y=\sqrt{8}=2\sqrt{2}$$:

   $$\frac{xy}{x+y}=\frac{\sqrt{2}\cdot 2\sqrt{2}}{\sqrt{2}+2\sqrt{2}}=\frac{4}{3\sqrt{2}}=\frac{2\sqrt{2}}{3}$$

   $$\frac{x-y}{xy}=\frac{\sqrt{2}-2\sqrt{2}}{\sqrt{2}\cdot 2\sqrt{2}}=\frac{-\sqrt{2}}{4}=-\frac{\sqrt{2}}{4}$$

2. б) При $$x=2-\sqrt{3},\ y=2+\sqrt{3}$$:

   $$xy=(2-\sqrt{3})(2+\sqrt{3})=4-3=1,$$
   $$x+y=(2-\sqrt{3})+(2+\sqrt{3})=4.$$

   $$\frac{xy}{x+y}=\frac{1}{4}$$

   $$x-y=(2-\sqrt{3})-(2+\sqrt{3})=-2\sqrt{3},$$
   $$\frac{x-y}{xy}=\frac{-2\sqrt{3}}{1}=-2\sqrt{3}$$

3. в) При $$x=\sqrt{6}-\sqrt{3},\ y=\sqrt{6}+\sqrt{3}$$:

   $$xy=(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})=6-3=3,$$
   $$x+y=(\sqrt{6}-\sqrt{3})+(\sqrt{6}+\sqrt{3})=2\sqrt{6}.$$

   $$\frac{xy}{x+y}=\frac{3}{2\sqrt{6}}=\frac{\sqrt{6}}{4}$$

   $$x-y=(\sqrt{6}-\sqrt{3})-(\sqrt{6}+\sqrt{3})=-2\sqrt{3},$$
   $$\frac{x-y}{xy}=\frac{-2\sqrt{3}}{3}=-\frac{2\sqrt{3}}{3}$$

4. г) При $$x=\sqrt{5}+\sqrt{2},\ y=\sqrt{5}-\sqrt{2}$$:

   $$xy=(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})=5-2=3,$$
   $$x+y=(\sqrt{5}+\sqrt{2})+(\sqrt{5}-\sqrt{2})=2\sqrt{5}.$$

   $$\frac{xy}{x+y}=\frac{3}{2\sqrt{5}}=\frac{3\sqrt{5}}{10}$$

   $$x-y=(\sqrt{5}+\sqrt{2})-(\sqrt{5}-\sqrt{2})=2\sqrt{2},$$
   $$\frac{x-y}{xy}=\frac{2\sqrt{2}}{3}$$

Ответ

а) $$\frac{2\sqrt{2}}{3},\ -\frac{\sqrt{2}}{4}$$;
б) $$\frac{1}{4},\ -2\sqrt{3}$$;
в) $$\frac{\sqrt{6}}{4},\ -\frac{2\sqrt{3}}{3}$$;
г) $$\frac{3\sqrt{5}}{10},\ \frac{2\sqrt{2}}{3}$$.