Задание 6 Избавьтесь от иррациональности: а) \frac{5}{\sqrt{2}}; \frac{1}{\sqrt{3}}; \frac{1}{\sqrt{5}} б) \frac{2}{\sqrt{2}+1};\frac{1}{\sqrt{11}-\sqrt{3}}; \frac{1}{3\sqrt{2}-1}

а) Избавимся от иррациональности:

1. \(\frac{5}{\sqrt{2}} = \frac{5 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{5\sqrt{2}}{2}\)
2. \(\frac{1}{\sqrt{3}} = \frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{3}}{3}\)
3. \(\frac{1}{\sqrt{5}} = \frac{1 \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{\sqrt{5}}{5}\)

б) Избавимся от иррациональности:

1. \(\frac{2}{\sqrt{2}+1} = \frac{2(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{2(\sqrt{2}-1)}{2-1} = 2(\sqrt{2}-1)\)
2. \(\frac{1}{\sqrt{11}-\sqrt{3}} = \frac{\sqrt{11}+\sqrt{3}}{(\sqrt{11}-\sqrt{3})(\sqrt{11}+\sqrt{3})} = \frac{\sqrt{11}+\sqrt{3}}{11-3} = \frac{\sqrt{11}+\sqrt{3}}{8}\)
3. \(\frac{1}{3\sqrt{2}-1} = \frac{3\sqrt{2}+1}{(3\sqrt{2}-1)(3\sqrt{2}+1)} = \frac{3\sqrt{2}+1}{(3\sqrt{2})^2-1^2} = \frac{3\sqrt{2}+1}{18-1} = \frac{3\sqrt{2}+1}{17}\)

Ответ: а) \(\frac{5\sqrt{2}}{2}\); \(\frac{\sqrt{3}}{3}\); \(\frac{\sqrt{5}}{5}\); б) \(2(\sqrt{2}-1)\); \(\frac{\sqrt{11}+\sqrt{3}}{8}\); \(\frac{3\sqrt{2}+1}{17}\)