ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 100

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\[1)\ \frac{2\sqrt{3}}{\sqrt{7}} = \frac{2\sqrt{3} \cdot \sqrt{7}}{\sqrt{7} \cdot \sqrt{7}} = \frac{2\sqrt{21}}{7}\]

\[2)\ \frac{4\sqrt{5}\ }{\sqrt{6}} = \frac{4\sqrt{5} \cdot \sqrt{6}}{\sqrt{6} \cdot \sqrt{6}} =\]

\[= \frac{4\sqrt{30}}{6} = \frac{2\sqrt{30}}{3}\]

\[3)\ \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}} =\]

\[= \frac{\left( \sqrt{7} + \sqrt{5} \right)\left( \sqrt{7} + \sqrt{5} \right)}{\left( \sqrt{7} - \sqrt{5} \right)\left( \sqrt{7} + \sqrt{5} \right)} =\]

\[= \frac{7 + 2\sqrt{35} + 5}{7 - 5} = \frac{12 + 2\sqrt{35}}{2} =\]

\[= 6 + \sqrt{35}\]

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

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

\[= \frac{3 + 2\sqrt{33} + 11}{3 - 11} =\]

\[= \frac{14 + 2\sqrt{33}}{- 8} =\]

\[= - \frac{7 + \sqrt{33}}{4}\]