Освободитесь от иррациональности в знаменателе дроби:
1) $\frac{\sqrt{2}}{\sqrt{2} + 1}$;
2) $\frac{4}{\sqrt{7} + \sqrt{3}}$;
3) $\frac{15}{\sqrt{15} - \sqrt{12}}$;
4) $\frac{19}{2\sqrt{5} - 1}$;
5) $\frac{1}{\sqrt{a} - \sqrt{b}}$;
6) $\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$.
$\frac{\sqrt{2}}{\sqrt{2} + 1} = \frac{\sqrt{2}(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{2 - \sqrt{2}}{(\sqrt{2})^2 - 1^2} = \frac{2 - \sqrt{2}}{2 - 1} = 2 - \sqrt{2}$
$\frac{4}{\sqrt{7} + \sqrt{3}} = \frac{4(\sqrt{7} - \sqrt{3})}{(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3})} = \frac{4(\sqrt{7} - \sqrt{3})}{(\sqrt{7})^2 - (\sqrt{3})^2} = \frac{4(\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{4(\sqrt{7} - \sqrt{3})}{4} = \sqrt{7} - \sqrt{3}$
$\frac{15}{\sqrt{15} - \sqrt{12}} = \frac{15(\sqrt{15} + \sqrt{12})}{(\sqrt{15} - \sqrt{12})(\sqrt{15} + \sqrt{12})} = \frac{15(\sqrt{15} + \sqrt{12})}{(\sqrt{15})^2 - (\sqrt{12})^2} = \frac{15(\sqrt{15} + \sqrt{12})}{15 - 12} = \frac{15(\sqrt{15} + \sqrt{12})}{3} = 5(\sqrt{15} + \sqrt{12}) = 5(\sqrt{5 * 3} + \sqrt{4 * 3}) = 5(\sqrt{5} * \sqrt{3} + 2\sqrt{3}) = 5\sqrt{3}(\sqrt{5} + 2)$
$\frac{19}{2\sqrt{5} - 1} = \frac{19(2\sqrt{5} + 1)}{(2\sqrt{5} - 1)(2\sqrt{5} + 1)} = \frac{19(2\sqrt{5} + 1)}{(2\sqrt{5})^2 - 1^2} = \frac{19(2\sqrt{5} + 1)}{4 * 5 - 1} = \frac{19(2\sqrt{5} + 1)}{20 - 1} = \frac{19(2\sqrt{5} + 1)}{19} = 2\sqrt{5} + 1$
$\frac{1}{\sqrt{a} - \sqrt{b}} = \frac{1 * (\sqrt{a} + \sqrt{b})}{(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})} = \frac{\sqrt{a} + \sqrt{b}}{(\sqrt{a})^2 - (\sqrt{b})^2} = \frac{\sqrt{a} + \sqrt{b}}{a - b}$
$\frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{(\sqrt{3})^2 + 2 * \sqrt{3} * 1 + 1^2}{3 - 1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = \frac{2(2 + \sqrt{3})}{2} = 2 + \sqrt{3}$
Пожауйста, оцените решение
