ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 503

 

\[\boxed{\text{503\ (503).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Разложим\ подкоренные\ \]

\[выражения\ на\ множителе\ и\ \]

\[вынесем\ общий\ множитель\ \]

\[за\ скобки.\]

\[\textbf{а)}\ \frac{\sqrt{70} - \sqrt{30}}{\sqrt{35} - \sqrt{15}} =\]

\[= \frac{\sqrt{7} \cdot \sqrt{10} - \sqrt{3} \cdot \sqrt{10}}{\sqrt{7} \cdot \sqrt{5} - \sqrt{3} \cdot \sqrt{5}} =\]

\[= \frac{\sqrt{10} \cdot \left( \sqrt{7} - \sqrt{3} \right)}{\sqrt{5} \cdot \left( \sqrt{7} - \sqrt{3} \right)} = \frac{\sqrt{2} \cdot \sqrt{5}}{\sqrt{5}} =\]

\[= \sqrt{2}\ \]

\[\textbf{б)}\ \frac{\sqrt{15} - 5}{\sqrt{6} - \sqrt{10}} = \frac{\sqrt{5}\sqrt{3} - \sqrt{5}\sqrt{5}}{\sqrt{3}\sqrt{2} - \sqrt{5}\sqrt{2}} =\]

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

\[\textbf{в)}\ \frac{2\sqrt{10} - 5}{4 - \sqrt{10}} = \frac{2\sqrt{5}\sqrt{2} - \sqrt{5}\sqrt{5}}{\sqrt{2}\sqrt{2^{3}} - \sqrt{2}\sqrt{5}} =\]

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

\[\textbf{г)}\ \frac{9 - 2\sqrt{3}}{3\sqrt{6} - 2\sqrt{2}} = \frac{\sqrt{3}\sqrt{3^{3}} - 2\sqrt{3}}{3\sqrt{2}\sqrt{3} - 2\sqrt{2}} =\]

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

\[\textbf{д)}\ \frac{2\sqrt{3} + 3\sqrt{2} - \sqrt{6}}{2 + \sqrt{6} - \sqrt{2}} =\]

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

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

\[= \sqrt{3}\]

\[\textbf{е)}\ \frac{\left( \sqrt{10} - 1 \right)^{2} - 3}{\sqrt{10} + \sqrt{3} - 1} =\]

\[= \frac{\left( \sqrt{10} - 1 - \sqrt{3} \right)\left( \sqrt{10} - 1 + \sqrt{3} \right)}{\sqrt{10} + \sqrt{3} - 1} =\]

\[= \sqrt{10} - 1 - \sqrt{3}\ \]

\[\boxed{\text{503.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

При всех допустимых значениях a верно равенство:

\[\left( \sqrt{a} \right)^{2} = a.\]

Решение.

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

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

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

\[= \frac{\sqrt{2} + \sqrt{3} - 1}{2 + 2\sqrt{6} + 3 - 1} =\]

\[= \frac{\sqrt{2} + \sqrt{3} - 1 \cdot \left( 4 + 2\sqrt{6} \right)}{4 - 2\sqrt{6} \cdot \left( 4 + 2\sqrt{6} \right)} =\]

\[= \frac{- 2\sqrt{2} + 2\sqrt{6} - 4}{- 8} =\]

\[= \frac{- 2 \cdot \left( 2 - \sqrt{6} + \sqrt{2} \right)}{- 8} =\]

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

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

\[= \frac{1 \cdot \left( \sqrt{5} - \sqrt{3} - 2 \right)}{\left( \sqrt{5} - \sqrt{3} + 2 \right) \cdot \left( \sqrt{5} - \sqrt{3} - 2 \right)} =\]

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

\[= \frac{\sqrt{5} - \sqrt{3} - 2}{5 - 2\sqrt{15} + 3 - 4} =\]

\[= \frac{\sqrt{5} - \sqrt{3} - 2 \cdot \left( 4 + 2\sqrt{15} \right)}{4 - 2\sqrt{15} \cdot \left( 4 + 2\sqrt{15} \right)} =\]

\[= \frac{6\sqrt{3} - 2\sqrt{5} - 8 - 4\sqrt{15}}{- 44} =\]

\[= \frac{4 + 2\sqrt{15} + \sqrt{5} - 3\sqrt{3}}{22}\ \]