ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 492

\[\boxed{\mathbf{492}.}\]

\[1)\ a^{\frac{1}{9}} \bullet \sqrt[6]{a^{3}\sqrt[3]{a}} =\]

\[= a^{\frac{1}{9}} \bullet \sqrt[6]{a^{3}} \bullet \sqrt[{6 \bullet 3}]{a} =\]

\[= a^{\frac{1}{9}} \bullet a^{\frac{3}{6}} \bullet a^{\frac{1}{18}} = a^{\frac{1}{9} + \frac{3}{6} + \frac{1}{18}} =\]

\[= a^{\frac{2 + 3 + 1}{18}} =\]

\[= a^{\frac{6}{18}} = a^{\frac{1}{3}};\]

\[2)\ b^{\frac{1}{12}} \bullet \sqrt[3]{b\sqrt[4]{b}} = b^{\frac{1}{12}} \bullet \sqrt[3]{b} \bullet\]

\[\bullet \sqrt[{3 \bullet 4}]{b} = b^{\frac{1}{12}} \bullet b^{\frac{1}{3}} \bullet b^{\frac{1}{12}} =\]

\[= b^{\frac{1}{12} + \frac{1}{3} + \frac{1}{12}} = b^{\frac{1 + 4 + 1}{12}} =\]

\[= b^{\frac{6}{12}} = b^{\frac{1}{2}};\]

\[3)\ \left( \sqrt[3]{a} + \sqrt[3]{b} \right) \bullet\]

\[\bullet \left( a^{\frac{2}{3}} + b^{\frac{2}{3}} - \sqrt[3]{\text{ab}} \right) =\]

\[= \left( a^{\frac{1}{3}} + b^{\frac{1}{3}} \right) \bullet\]

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

\[= \left( a^{\frac{1}{3}} \right)^{3} + \left( b^{\frac{1}{3}}\ \right)^{3} = a + b.\]