Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 722

\[1)\ \frac{5}{\sqrt{3} - \sqrt{2}} =\]

\[= \frac{5\left( \sqrt{3} + \sqrt{2} \right)}{\left( \sqrt{3} - \sqrt{2} \right)\left( \sqrt{3} + \sqrt{2} \right)} =\]

\[= \frac{5\left( \sqrt{3} + \sqrt{2} \right)}{3 - 2} = 5\left( \sqrt{3} + \sqrt{2} \right);\]

\[2)\ \frac{3}{\sqrt{6} + \sqrt{5}} =\]

\[= \frac{3\left( \sqrt{6} - \sqrt{5} \right)}{\left( \sqrt{6} + \sqrt{5} \right)\left( \sqrt{6} - \sqrt{5} \right)} =\]

\[= \frac{3\left( \sqrt{6} - \sqrt{5} \right)}{6 - 5} = 3\left( \sqrt{6} - \sqrt{5} \right);\]

\[3)\ \frac{12}{\sqrt{10} - \sqrt{7}} =\]

\[= \frac{12\left( \sqrt{10} + \sqrt{7} \right)}{\left( \sqrt{10} - \sqrt{7} \right)\left( \sqrt{10} + \sqrt{7} \right)} =\]

\[= \frac{12\left( \sqrt{10} + \sqrt{7} \right)}{10 - 7} = 4\left( \sqrt{10} + \sqrt{7} \right);\]

\[4)\ \frac{8}{\sqrt{11} + \sqrt{3}} =\]

\[= \frac{8\left( \sqrt{11} - \sqrt{3} \right)}{\left( \sqrt{11} + \sqrt{3} \right)\left( \sqrt{11} - \sqrt{3} \right)} =\]

\[= \frac{8\left( \sqrt{11} - \sqrt{3} \right)}{11 - 3} = \sqrt{11} - \sqrt{3}.\]