Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 112

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Год:2020-2021-2022-2023
Тип:учебник
Серия:Базовый и углубленный уровни

Задание 112

\[\boxed{\mathbf{112}\mathbf{.}}\]

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

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

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

\[2)\ \frac{\sqrt{5}}{5 + \sqrt{10}} =\]

\[= \frac{\sqrt{5} \bullet \left( 5 - \sqrt{10} \right)}{\left( 5 + \sqrt{10} \right) \bullet \left( 5 - \sqrt{10} \right)} =\]

\[= \frac{\sqrt{5}\left( 5 - \sqrt{10} \right)}{25 - 10} =\]

\[= \frac{\sqrt{5}\left( 5 - \sqrt{10} \right)}{15} =\]

\[= \frac{5\sqrt{5} - \sqrt{5 \bullet 10}}{15} =\]

\[= \frac{5\sqrt{5} - \sqrt{25 \bullet 2}}{15} = \frac{5\sqrt{5} - 5\sqrt{2}}{15} =\]

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

\[3)\ \frac{3}{\sqrt[3]{4}} = \frac{3 \bullet \sqrt[3]{2}}{\sqrt[3]{2^{2}} \bullet \sqrt[3]{2}} = \frac{3\sqrt[3]{2}}{\sqrt[3]{2^{2} \bullet 2}} =\]

\[= \frac{3\sqrt[3]{2}}{\sqrt[3]{2^{3}}} = \frac{3\sqrt[3]{2}}{2}\]

\[4)\ \frac{2}{\sqrt[4]{27}} = \frac{2 \bullet \sqrt[4]{3}}{\sqrt[4]{3^{3}} \bullet \sqrt[4]{3}} = \frac{2\sqrt[4]{3}}{\sqrt[4]{3^{3} \bullet 3}} =\]

\[= \frac{2\sqrt[4]{3}}{\sqrt[4]{3^{4}}} = \frac{2\sqrt[4]{3}}{3}\]

\[5)\ \frac{3}{\sqrt[4]{5} - \sqrt[4]{2}} =\]

\[= \frac{3 \bullet \left( \sqrt[4]{5} + \sqrt[4]{2} \right)}{\left( \sqrt[4]{5} - \sqrt[4]{2} \right) \bullet \left( \sqrt[4]{5} + \sqrt[4]{2} \right)} =\]

\[= \frac{3\left( \sqrt[4]{5} + \sqrt[4]{2} \right)}{\sqrt{5} - \sqrt{2}} =\]

\[= \frac{3\left( \sqrt[4]{5} + \sqrt[4]{2} \right) \bullet \left( \sqrt{5} + \sqrt{2} \right)}{\left( \sqrt{5} - \sqrt{2} \right) \bullet \left( \sqrt{5} + \sqrt{2} \right)} =\]

\[= \frac{3\left( \sqrt[4]{5} + \sqrt[4]{2} \right)\left( \sqrt{5} + \sqrt{2} \right)}{5 - 2} =\]

\[= \left( \sqrt[4]{5} + \sqrt[4]{2} \right)\left( \sqrt{5} + \sqrt{2} \right)\]

\[6)\ \frac{11}{\sqrt[3]{3} + \sqrt[3]{2}} = \frac{11}{3^{\frac{1}{3}} + 2^{\frac{1}{3}}} =\]

\[= \frac{11 \bullet \left( 3^{\frac{2}{3}} - 3^{\frac{1}{3}}2^{\frac{1}{3}} + 2^{\frac{2}{3}} \right)}{\left( 3^{\frac{1}{3}} + 2^{\frac{1}{3}} \right) \bullet \left( 3^{\frac{2}{3}} - 3^{\frac{1}{3}} \bullet 2^{\frac{1}{3}} + 2^{\frac{2}{3}} \right)} =\]

\[= \frac{11 \bullet \left( \sqrt[3]{3^{2}} - \sqrt[3]{3 \bullet 2} + \sqrt[3]{2^{2}} \right)}{\left( 3^{\frac{1}{3}} \right)^{3} + \left( 2^{\frac{1}{3}} \right)^{3}} =\]

\[= \frac{11\left( \sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4} \right)}{3 + 2} =\]

\[= \frac{11\left( \sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4} \right)}{5}\]

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

\[= \frac{1 + \sqrt{2} - \sqrt{3}}{\left( 1 + \sqrt{2} + \sqrt{3} \right)\left( 1 + \sqrt{2} - \sqrt{3} \right)} =\]

\[= \frac{1 + \sqrt{2} - \sqrt{3}}{\left( 1 + \sqrt{2} \right)^{2} - \left( \sqrt{3} \right)^{2}} =\]

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

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

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

\[= \frac{\sqrt{2} + 2 - \sqrt{6}}{4}\]

\[8)\ \frac{1}{\sqrt[3]{4} + \sqrt[3]{6} + \sqrt[3]{9}} =\]

\[= \frac{1}{4^{\frac{1}{3}} + 6^{\frac{1}{3}} + 9^{\frac{1}{3}}} =\]

\[= \frac{\sqrt[3]{2} - \sqrt[3]{3}}{\left( 2^{\frac{1}{3}} \right)^{3} - \left( 3^{\frac{1}{3}} \right)^{3}} = \frac{\sqrt[3]{2} - \sqrt[3]{3}}{2 - 3} =\]

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

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