Упр.416(стар учебник) ГДЗ Дорофеев Суворова 8 класс (Алгебра)

Задача

а) v3/(3-v6)-v2/(3+v6);
б) v7+v7/(2+v7) ((2+v7)/(2-v7)-2-v7);
в) ((v5-5v2)/(v5+v2)+v10)(v5/v2+v2/v5);
г) (3v3-v5)/(3v3+v5)-(3v3+v5)/(3v3-v5).

Подробный ответ

а)
$$\frac{\sqrt{3}}{3-\sqrt{6}}-\frac{\sqrt{2}}{3+\sqrt{6}}= \frac{\sqrt{3}(3+\sqrt{6})-\sqrt{2}(3-\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}$$
$$=\frac{3\sqrt{3}+3\sqrt{2}-3\sqrt{2}+2\sqrt{3}}{9-6} =\frac{5\sqrt{3}}{3}$$
б)
$$\sqrt{7}+\frac{\sqrt{7}}{2+\sqrt{7}}\left(\frac{2+\sqrt{7}}{2-\sqrt{7}}-2-\sqrt{7}\right)$$
$$=\sqrt{7}+\frac{\sqrt{7}}{2+\sqrt{7}}\cdot \frac{2+\sqrt{7}-(2+\sqrt{7})(2-\sqrt{7})}{2-\sqrt{7}}$$
$$=\sqrt{7}+\frac{\sqrt{7}}{2+\sqrt{7}}\cdot \frac{2+\sqrt{7}-4+7}{2-\sqrt{7}} =\sqrt{7}+\frac{\sqrt{7}(5+\sqrt{7})}{(2+\sqrt{7})(2-\sqrt{7})}$$
$$=\sqrt{7}+\frac{5\sqrt{7}+7}{4-7} =\sqrt{7}-\frac{5\sqrt{7}+7}{3} =\frac{3\sqrt{7}-5\sqrt{7}-7}{3}$$
$$=-\frac{2\sqrt{7}+7}{3}$$
в)
$$\left(\frac{\sqrt{5}-5\sqrt{2}}{\sqrt{5}+\sqrt{2}}+\sqrt{10}\right) \left(\frac{\sqrt{5}}{\sqrt{2}}+\frac{\sqrt{2}}{\sqrt{5}}\right)$$
$$=\left(\frac{\sqrt{5}-5\sqrt{2}+\sqrt{10}(\sqrt{5}+\sqrt{2})}{\sqrt{5}+\sqrt{2}}\right) \left(\frac{(\sqrt{5})^2+(\sqrt{2})^2}{\sqrt{2}\cdot\sqrt{5}}\right)$$
$$=\frac{\sqrt{5}-5\sqrt{2}+\sqrt{50}+\sqrt{20}}{\sqrt{5}+\sqrt{2}}\cdot\frac{7}{\sqrt{10}}$$
$$=\frac{\sqrt{5}-5\sqrt{2}+5\sqrt{2}+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}\cdot\frac{7}{\sqrt{10}} =\frac{3\sqrt{5}}{\sqrt{5}+\sqrt{2}}\cdot\frac{7}{\sqrt{10}}$$
$$=\frac{21\sqrt{5}}{\sqrt{10}(\sqrt{5}+\sqrt{2})} =\frac{21}{\sqrt{2}(\sqrt{5}+\sqrt{2})} =\frac{21}{\sqrt{10}+2}$$
г)
$$\frac{3\sqrt{3}-\sqrt{5}}{3\sqrt{3}+\sqrt{5}}-\frac{3\sqrt{3}+\sqrt{5}}{3\sqrt{3}-\sqrt{5}}$$
$$=\frac{(3\sqrt{3}-\sqrt{5})^2-(3\sqrt{3}+\sqrt{5})^2}{(3\sqrt{3}+\sqrt{5})(3\sqrt{3}-\sqrt{5})}$$
$$=\frac{27-6\sqrt{15}+5-27-6\sqrt{15}-5}{27-5} =\frac{-12\sqrt{15}}{22}$$
$$=-\frac{6\sqrt{15}}{11}$$