ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 40

\[\boxed{\mathbf{40}\mathbf{.}}\]

\[1)\ \sqrt[4]{324}\ :\sqrt[4]{4} = \sqrt[4]{\frac{324}{4}} = \sqrt[4]{81} =\]

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

\[2)\ \sqrt[3]{128}\ :\ \sqrt[3]{2000} = \sqrt[3]{\frac{128}{2000}} =\]

\[= \sqrt[3]{\frac{64}{1000}} = \frac{\sqrt[3]{64}}{\sqrt[3]{1000}} = \frac{\sqrt[3]{4^{3}}}{\sqrt[3]{10^{3}}} =\]

\[= \frac{4}{10} = 0,4\]

\[3)\ \frac{\sqrt[3]{16}}{\sqrt[3]{2}} = \sqrt[3]{\frac{16}{2}} = \sqrt[3]{8} = \sqrt[3]{2^{3}} = 2\]

\[4)\ \frac{\sqrt[5]{256}}{\sqrt[5]{8}} = \sqrt[5]{\frac{256}{8}} = \sqrt[5]{32} =\]

\[= \sqrt[5]{2^{5}} = 2\]

\[5)\ \left( \sqrt{25} - \sqrt{45} \right)\ :\sqrt{5} =\]

\[= \frac{\sqrt{25}}{\sqrt{5}} - \frac{\sqrt{45}}{\sqrt{5}} = \sqrt{\frac{25}{5}} - \sqrt{\frac{45}{5}} =\]

\[= \sqrt{5} - \sqrt{9} = \sqrt{5} - 3\]

\[6)\ \left( \sqrt[3]{625} - \sqrt[3]{5} \right)\ :\sqrt[3]{5} =\]

\[= \frac{\sqrt[3]{625}}{\sqrt[3]{5}} - \frac{\sqrt[3]{5}}{\sqrt[3]{5}} = \sqrt[3]{\frac{625}{5}} - 1 =\]

\[= \sqrt[3]{125} - 1 = \sqrt[3]{5^{3}} - 1 =\]

\[= 5 - 1 = 4\]