Осовободитесь от иррациональности в знаменателе дроби:
1) $\frac{a}{\sqrt{11}}$;
2) $\frac{18}{\sqrt{6}}$;
3) $\frac{5}{\sqrt{10}}$;
4) $\frac{13}{\sqrt{26}}$;
5) $\frac{30}{\sqrt{15}}$;
6) $\frac{2}{3\sqrt{x}}$.
$\frac{a}{\sqrt{11}} = \frac{a * \sqrt{11}}{\sqrt{11} * \sqrt{11}} = \frac{a\sqrt{11}}{(\sqrt{11})^2} = \frac{a\sqrt{11}}{11}$
$\frac{18}{\sqrt{6}} = \frac{18 * \sqrt{6}}{\sqrt{6} * \sqrt{6}} = \frac{18\sqrt{6}}{(\sqrt{6})^2} = \frac{18\sqrt{6}}{6} = 3\sqrt{6}$
$\frac{5}{\sqrt{10}} = \frac{5 * \sqrt{10}}{\sqrt{10} * \sqrt{10}} = \frac{5\sqrt{10}}{(\sqrt{10})^2} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2}$
$\frac{13}{\sqrt{26}} = \frac{13 * \sqrt{26}}{\sqrt{26} * \sqrt{26}} = \frac{13\sqrt{26}}{(\sqrt{26})^2} = \frac{13\sqrt{26}}{26} = \frac{\sqrt{26}}{2}$
$\frac{30}{\sqrt{15}} = \frac{30 * \sqrt{15}}{\sqrt{15} * \sqrt{15}} = \frac{30\sqrt{15}}{(\sqrt{15})^2} = \frac{30\sqrt{15}}{15} = 2\sqrt{15}$
$\frac{2}{3\sqrt{x}} = \frac{2 * \sqrt{x}}{3\sqrt{x} * \sqrt{x}} = \frac{2\sqrt{x}}{3(\sqrt{x})^2} = \frac{2\sqrt{x}}{3x}$
Пожауйста, оцените решение