Страница 38 номер 112, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Освободиться от иррациональности в знаменателе дроби: 1) \(\frac{2}{\sqrt{2} - \sqrt{3}}\); 2) \(\frac{\sqrt{5}}{5 + \sqrt{10}}\); 3) \(\frac{3}{\sqrt[3]{4}}\); 4) \(\frac{2}{\sqrt[4]{27}}\); 5) \(\frac{3}{\sqrt[4]{5} - \sqrt[4]{2}}\); 6) \(\frac{11}{\sqrt[3]{3} + \sqrt[3]{2}}\); 7) \(\frac{1}{1 + \sqrt{2} + \sqrt{3}}\); 8) \(\frac{1}{\sqrt[3]{4} + \sqrt[3]{6} + \sqrt[3]{9}}\).
1) \(\frac{2}{\sqrt{2} - \sqrt{3}} = \frac{2\left(\sqrt{2} + \sqrt{3}\right)}{\left(\sqrt{2} - \sqrt{3}\right)\left(\sqrt{2} + \sqrt{3}\right)} = \frac{2\left(\sqrt{2} + \sqrt{3}\right)}{2 - 3} = -2\left(\sqrt{2} + \sqrt{3}\right);\)
2) \(\frac{\sqrt{5}}{5 + \sqrt{10}} = \frac{\sqrt{5} \cdot \left(5 - \sqrt{10}\right)}{\left(5 + \sqrt{10}\right)\left(5 - \sqrt{10}\right)} = \frac{5\sqrt{5} - \sqrt{50}}{25 - 10} = \frac{5\left(\sqrt{5} - \sqrt{2}\right)}{15} = \frac{\sqrt{5} - \sqrt{2}}{3};\)
3) \(\frac{3}{\sqrt[3]{4}} = \frac{3\sqrt[3]{2}}{\sqrt[3]{4} \cdot \sqrt[3]{2}} = \frac{3\sqrt[3]{2}}{\sqrt[3]{2^{3}}} = \frac{3\sqrt[3]{2}}{2};\)
4) \(\frac{2}{\sqrt[4]{27}} = \frac{2\sqrt[4]{3}}{\sqrt[4]{3^{3}} \cdot \sqrt[4]{3}} = \frac{2}{3}\sqrt[4]{3};\)
5) \(\frac{3}{\sqrt[4]{5} - \sqrt[4]{2}} = \frac{3\left(\sqrt[4]{5} + \sqrt[4]{2}\right)}{\sqrt{5} - \sqrt{2}} = \frac{3(\sqrt[4]{5} + \sqrt[4]{2})(\sqrt{5} + \sqrt{2})}{5 - 2} = (\sqrt[4]{5} + \sqrt[4]{2})(\sqrt{5} + \sqrt{2});\)
6) \(\frac{11}{\sqrt[3]{3} + \sqrt[3]{2}} = \frac{11\left(\sqrt[3]{3^{2}} - \sqrt[3]{3}\sqrt[3]{2} + \sqrt[3]{2^{2}}\right)}{\left(\sqrt[3]{3} + \sqrt[3]{2}\right)\left(\sqrt[3]{3^{2}} - \sqrt[3]{3}\sqrt[3]{2} + \sqrt[3]{2^{2}}\right)} = \frac{11\left(\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}\right)}{3 + 2} =\)
\(= \dfrac{11}{5}\left(\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}\right);\)
7) \(\frac{1}{1 + \sqrt{2} + \sqrt{3}} = \frac{\left(1 + \sqrt{2}\right) - \sqrt{3}}{\left(\left(1 + \sqrt{2}\right) + \sqrt{3}\right)\left(\left(1 + \sqrt{2}\right) - \sqrt{3}\right)} =\)
\(= \frac{1 + \sqrt{2} - \sqrt{3}}{\left(1 + \sqrt{2}\right)^{2} - 3} = \frac{1 + \sqrt{2} - \sqrt{3}}{1 + 2\sqrt{2} + 2 - 3} = \frac{1 + \sqrt{2} - \sqrt{3}}{2\sqrt{2}} =\)
\(= \dfrac{\left(1 + \sqrt{2} - \sqrt{3}\right)\sqrt{2}}{2\sqrt{2} \cdot \sqrt{2}} = \dfrac{\sqrt{2} + 2 - \sqrt{6}}{4};\)
8) \(\frac{1}{\sqrt[3]{4} + \sqrt[3]{6} + \sqrt[3]{9}} = \frac{\left(\sqrt[3]{3} - \sqrt[3]{2}\right)}{\left(\sqrt[3]{3^{2}} + \sqrt[3]{3 \cdot 2} + \sqrt[3]{2^{2}}\right)\left(\sqrt[3]{3} - \sqrt[3]{2}\right)} = \frac{\sqrt[3]{3} - \sqrt[3]{2}}{3 - 2} = \sqrt[3]{3} - \sqrt[3]{2}.\)
