Освободитесь от иррациональности в знаменателе дроби:
1) $\frac{1}{\sqrt{6} + \sqrt{2} + 1}$;
2) $\frac{2}{\sqrt{10} + \sqrt{5} - \sqrt{3}}$.
$\frac{1}{\sqrt{6} + \sqrt{2} + 1} = \frac{\sqrt{6} + \sqrt{2} - 1}{(\sqrt{6} + \sqrt{2} + 1)(\sqrt{6} + \sqrt{2} - 1)} = \frac{\sqrt{6} + \sqrt{2} - 1}{(\sqrt{6} + \sqrt{2})^2 - 1^2} = \frac{\sqrt{6} + \sqrt{2} - 1}{6 + 2\sqrt{12} + 2 - 1} = \frac{\sqrt{6} + \sqrt{2} - 1}{7 + 2\sqrt{12}} = \frac{(\sqrt{6} + \sqrt{2} - 1)(7 - 2\sqrt{12})}{(7 + 2\sqrt{12})(7 - 2\sqrt{12})} = \frac{\sqrt{6} + \sqrt{2} - 1}{7 + 2\sqrt{12}} = \frac{(\sqrt{6} + \sqrt{2} - 1)(7 - 2\sqrt{12})}{7^2 - (2\sqrt{12})^2} = \frac{7\sqrt{6} + 7\sqrt{2} - 7 - 2\sqrt{72} - 2\sqrt{24} + 2\sqrt{12}}{7^2 - (2\sqrt{12})^2} = \frac{7\sqrt{6} + 7\sqrt{2} - 7 - 2\sqrt{36 * 2} - 2\sqrt{4 * 6} + 2\sqrt{4 * 3}}{49 - 4 * 12} = \frac{7\sqrt{6} + 7\sqrt{2} - 7 - 12\sqrt{2} - 4\sqrt{6} + 4\sqrt{3}}{49 - 48} = 3\sqrt{6} - 5\sqrt{2} + 4\sqrt{3} - 7$
$\frac{2}{\sqrt{10} + \sqrt{5} - \sqrt{3}} = \frac{2(\sqrt{10} + \sqrt{5} + \sqrt{3})}{(\sqrt{10} + \sqrt{5} - \sqrt{3})(\sqrt{10} + \sqrt{5} + \sqrt{3})} = \frac{2(\sqrt{10} + \sqrt{5} + \sqrt{3})}{(\sqrt{10} + \sqrt{5})^2 - (\sqrt{3})^2} = \frac{2(\sqrt{10} + \sqrt{5} + \sqrt{3})}{10 + 2\sqrt{50} + 5 - 3} = \frac{2(\sqrt{10} + \sqrt{5} + \sqrt{3})}{12 + 2\sqrt{25 * 2}} = \frac{2(\sqrt{10} + \sqrt{5} + \sqrt{3})}{12 + 10\sqrt{2}} = \frac{2(\sqrt{10} + \sqrt{5} + \sqrt{3})}{2(6 + 5\sqrt{2})} = \frac{(\sqrt{10} + \sqrt{5} + \sqrt{3})(6 - 5\sqrt{2})}{(6 + 5\sqrt{2})(6 - 5\sqrt{2})} = \frac{6\sqrt{10} + 6\sqrt{5} + 6\sqrt{3} - 5\sqrt{20} - 5\sqrt{10} - 5\sqrt{6}}{6^2 - (5\sqrt{2})^2} = \frac{6\sqrt{10} + 6\sqrt{5} + 6\sqrt{3} - 5\sqrt{4 * 5} - 5\sqrt{10} - 5\sqrt{6}}{36 - 25 * 2} = \frac{6\sqrt{10} + 6\sqrt{5} + 6\sqrt{3} - 10\sqrt{5} - 5\sqrt{10} - 5\sqrt{6}}{36 - 50} = \frac{\sqrt{10} - 4\sqrt{5} + 6\sqrt{3} - 5\sqrt{6}}{-14} = -\frac{\sqrt{10} - 4\sqrt{5} + 6\sqrt{3} - 5\sqrt{6}}{14}$
Пожауйста, оцените решение