Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 58

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\[1)\ 2^{\frac{4}{5}} \bullet 2^{\frac{11}{5}} = 2^{\frac{4 + 11}{5}} = 2^{\frac{15}{5}} =\]

\[= 2^{3} = 8\]

\[2)\ 5^{\frac{2}{7}} \bullet 5^{\frac{5}{7}} = 5^{\frac{2 + 5}{7}} = 5^{\frac{7}{7}} = 5^{1} = 5\]

\[3)\ 9^{\frac{2}{3}}\ :9^{\frac{1}{6}} = 9^{\frac{2}{3} - \frac{1}{6}} = 9^{\frac{4 - 1}{6}} = 9^{\frac{3}{6}} =\]

\[= 9^{\frac{1}{2}} = \left( 3^{2} \right)^{\frac{1}{2}} = 3\]

\[4)\ 4^{\frac{1}{3}}\ :4^{\frac{5}{6}} = 4^{\frac{1}{3} - \frac{5}{6}} = 4^{\frac{2 - 5}{6}} =\]

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

\[5)\ \left( 8^{\frac{1}{12}} \right)^{- 4} = 8^{- \frac{4}{12}} = 8^{- \frac{1}{3}} =\]

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