Решебник по алгебре 9 класс Макарычев Задание 197

\[\boxed{\text{197\ (197).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{3 + 3^{\frac{1}{2}}}{3^{- \frac{1}{2}}} = \frac{3 + \sqrt{3}}{\frac{1}{\sqrt{3}}} = 3\sqrt{3} + 3;\]

\[\textbf{б)}\ \frac{10}{10 - 10^{\frac{1}{2}}} = \frac{10}{10 - \sqrt{10}} =\]

\[= \frac{10}{\sqrt{10} \cdot (\sqrt{10} - 1)} = \frac{\sqrt{10}}{\sqrt{10} - 1};\]

\[\textbf{в)}\ \frac{x - y}{x^{\frac{1}{2}} + y^{\frac{1}{2}}} =\]

\[= \frac{\left( \sqrt{x} - \sqrt{y} \right)\left( \sqrt{x} + \sqrt{y} \right)}{\sqrt{x} + \sqrt{y}} =\]

\[= \sqrt{x} - \sqrt{y}\]

\[\textbf{г)}\ \frac{b^{\frac{1}{2}} - 5}{b - 25} =\]

\[= \frac{\sqrt{b} - 5}{(\sqrt{b} - 5)(\sqrt{b} + 5)} = \frac{1}{\sqrt{b} + 5}\]

\[\textbf{д)}\ \frac{c + 2c^{\frac{1}{2}}d^{\frac{1}{2}} + d}{c - d} =\]

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

\[= \frac{\sqrt{c} + \sqrt{d}}{\sqrt{c} - \sqrt{d}}\]

\[\textbf{е)}\ \frac{m + n}{m^{\frac{2}{3}} - m^{\frac{1}{3}}n^{\frac{1}{3}} + n^{\frac{2}{3}}} =\]

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

\[= m^{\frac{1}{3}} + n^{\frac{1}{3}}\ \]

\[\boxed{\mathbf{197}\text{.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]