Упр.2.153 ГДЗ Дорофеев Суворова 8 класс (Алгебра)

Упростите выражение:
а) $$\frac{1}{\sqrt{6}+\sqrt{5}}+\frac{1}{\sqrt{6}-\sqrt{5}}$$;
б) $$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}-\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$.

а)

$$\frac{1}{\sqrt{6}+\sqrt{5}}+\frac{1}{\sqrt{6}-\sqrt{5}} = \frac{\sqrt{6}-\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{6}-\sqrt{5})} + \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})}$$

$$= \frac{\sqrt{6}-\sqrt{5}+\sqrt{6}+\sqrt{5}}{6-5} = \frac{2\sqrt{6}}{1} = 2\sqrt{6}$$

б)

$$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}-\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{(\sqrt{3}-\sqrt{2})^2-(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$$

$$= \frac{(3-2\sqrt{6}+2)-(3+2\sqrt{6}+2)}{3-2} = \frac{5-2\sqrt{6}-5-2\sqrt{6}}{1} = -4\sqrt{6}$$

Ответ

а) $$2\sqrt{6}$$; б) $$-4\sqrt{6}$$.