ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 473

\[\boxed{\mathbf{473}.}\]

\[1)\ \sqrt[3]{a} \bullet \sqrt[6]{a} = a^{\frac{1}{3}} \bullet a^{\frac{1}{6}} = a^{\frac{1}{3} + \frac{1}{6}} =\]

\[= a^{\frac{2 + 1}{6}} = a^{\frac{3}{6}} = a^{\frac{1}{2}} = \sqrt{a};\]

\[Если\ a = 0,16:\]

\[\sqrt{a} = \sqrt{0,16} = 0,4.\]

\[2)\ \sqrt{b}\ :\sqrt[6]{b} = b^{\frac{1}{2}}\ :b^{\frac{1}{6}} = b^{\frac{1}{2} - \frac{1}{6}} =\]

\[= b^{\frac{3 - 1}{6}} = b^{\frac{2}{6}} = b^{\frac{1}{3}} = \sqrt[3]{b};\]

\[Если\ b = 0,027:\]

\[\sqrt[3]{b} = \sqrt[3]{0,027} = 0,3.\]

\[3)\ \frac{\sqrt{x}\ :\sqrt[3]{x^{2}}}{\sqrt[6]{x}} = \frac{x^{\frac{1}{2}}\ :x^{\frac{2}{3}}}{x^{\frac{1}{6}}} =\]

\[= x^{\frac{1}{2} - \frac{2}{3} - \frac{1}{6}} = x^{\frac{3 - 4 - 1}{6}} = x^{- \frac{2}{6}} = x^{- \frac{1}{3}};\]

\[Если\ x = 1,33:\]

\[(1,33)^{- \frac{1}{3}} = \frac{1}{\sqrt[3]{1,33}}.\]

\[4)\ \sqrt[3]{a} \bullet \sqrt[4]{a} \bullet \sqrt[12]{a^{5}} =\]

\[= a^{\frac{1}{3}} \bullet a^{\frac{1}{4}} \bullet a^{\frac{5}{12}} = a^{\frac{1}{3} + \frac{1}{4} + \frac{5}{12}} =\]

\[= a^{\frac{4 + 3 + 5}{12}} = a^{\frac{12}{12}} = a;\]

\[Если\ a = 3,75:\]

\[{3,75}^{1} = 3,75.\]