Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 95

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\[1)\ \sqrt[3]{5^{3} \bullet 7^{3}} = \sqrt[3]{5^{3}} \bullet \sqrt[3]{7^{3}} =\]

\[= 5 \bullet 7 = 35;\]

\[\sqrt[4]{324} \bullet \sqrt[4]{4} = \sqrt[4]{324 \bullet 4} =\]

\[= \sqrt[4]{81 \bullet 4 \bullet 4} = \sqrt[4]{81 \bullet 16} =\]

\[= \sqrt[4]{3^{4} \bullet 2^{4}} = 3 \bullet 2 = 6;\]

\[\sqrt[4]{15\frac{5}{8}}\ :\sqrt[4]{\frac{2}{5}} = \sqrt[4]{\frac{15 \bullet 8 + 5}{8}} \bullet \sqrt[4]{\frac{5}{2}} =\]

\[= \sqrt[4]{\frac{125}{8} \bullet \frac{5}{2}} = \sqrt[4]{\frac{5^{3} \bullet 5}{2^{3} \bullet 2}} = \sqrt[4]{\frac{5^{4}}{2^{4}}} =\]

\[= \frac{5}{2} = 2,5.\]

\[2)\ 56^{0}\ :8^{- 2} = 1\ :\left( \frac{1}{8} \right)^{2} =\]

\[= 1 \bullet 8^{2} = 64;\]

\[16^{\frac{1}{4}} \bullet 25^{\frac{1}{2}} = \left( 2^{4} \right)^{\frac{1}{4}} \bullet \left( 5^{2} \right)^{\frac{1}{2}} =\]

\[= 2 \bullet 5 = 10;\]

\[\left( \frac{1}{15} \right)^{- 1}\ :9^{\frac{1}{2}} = 15\ :\left( 3^{2} \right)^{\frac{1}{2}} =\]

\[= 15\ :3 = 5;\]

\[8^{\frac{1}{3}} \bullet \left( \frac{1}{2} \right)^{4}\ :16^{- 1} =\]

\[= \left( 2^{3} \right)^{\frac{1}{3}} \bullet \frac{1}{2^{4}}\ :\frac{1}{16} = 2 \bullet \frac{1}{16} \bullet 16 =\]

\[= 2.\]

\[3)\ \frac{5^{\frac{1}{4}} \bullet 5^{- \frac{1}{4}}}{5^{2}} = 5^{\frac{1}{4} + \left( - \frac{1}{4} \right) - 2} = 5^{- 2} =\]

\[= \frac{1}{5^{2}} = \frac{1}{25};\]

\[\frac{7^{\frac{7}{3}} \bullet 7^{- \frac{4}{3}}}{7^{2}} = 7^{\frac{7}{3} + \left( - \frac{4}{3} \right) - 2} = 7^{\frac{3}{3} - 2} =\]

\[= 7^{1 - 2} = 7^{- 1} = \frac{1}{7};\ \]

\[\frac{{0,3}^{0,3} \bullet {0,3}^{- 1}}{{0,3}^{1,3}} = {0,3}^{0,3 + ( - 1) - 1,3} =\]

\[= {0,3}^{- 2} = \left( \frac{3}{10} \right)^{- 2} = \left( \frac{10}{3} \right)^{2} =\]

\[= \frac{100}{9} = 11\frac{1}{9}.\]