Упр.7.40 ГДЗ Мордковича 11 класс профильный уровень (Алгебра)

Освободимся от иррациональности в знаменателе.

а)

$$\frac{3}{\sqrt{15}+\sqrt{10}-\sqrt{6}-2} = \frac{3}{\sqrt{5}(\sqrt{3}+\sqrt{2})-\sqrt{2}(\sqrt{3}+\sqrt{2})} = \frac{3}{(\sqrt{5}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$$
$$= \frac{3(\sqrt{5}+\sqrt{2})(\sqrt{3}-\sqrt{2})}{(5-2)(3-2)} = (\sqrt{5}+\sqrt{2})(\sqrt{3}-\sqrt{2})$$
$$=\sqrt{15}-\sqrt{10}+\sqrt{6}-2.$$

б)

$$\frac{6}{\sqrt{10}-\sqrt{6}+5-\sqrt{15}} = \frac{6}{\sqrt{2}(\sqrt{5}-\sqrt{3})+\sqrt{5}(\sqrt{5}-\sqrt{3})} = \frac{6}{(\sqrt{2}+\sqrt{5})(\sqrt{5}-\sqrt{3})}$$
$$= \frac{6(\sqrt{2}-\sqrt{5})(\sqrt{5}+\sqrt{3})}{(2-5)(5-3)} = (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{3})$$
$$=5+\sqrt{15}-\sqrt{10}-\sqrt{6}.$$

Ответ: а) $$\sqrt{15}-\sqrt{10}+\sqrt{6}-2$$; б) $$5+\sqrt{15}-\sqrt{10}-\sqrt{6}$$.