Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 97

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

\[= \sqrt[3]{\frac{3}{2}} \bullet \sqrt[3]{\frac{2 \bullet 4 + 1}{4}} = \sqrt[3]{\frac{3}{2} \bullet \frac{9}{4}} =\]

\[= \sqrt[3]{\frac{3 \bullet 3^{2}}{2 \bullet 2^{2}}} = \sqrt[3]{\frac{3^{3}}{2^{3}}} = \frac{3}{2} = 1,5\]

\[2)\ \sqrt[4]{\frac{3}{4}} \bullet \sqrt[4]{6\frac{3}{4}} =\]

\[= \sqrt[4]{\frac{3}{4}} \bullet \sqrt[4]{\frac{6 \bullet 4 + 3}{4}} = \sqrt[4]{\frac{3}{4} \bullet \frac{27}{4}} =\]

\[= \sqrt[4]{\frac{3 \bullet 3^{3}}{2^{2} \bullet 2^{2}}} = \sqrt[4]{\frac{3^{4}}{2^{4}}} = \frac{3}{2} = 1,5\]

\[3)\ \sqrt[4]{15\frac{5}{8}}\ :\sqrt[4]{\frac{2}{5}} =\]

\[= \sqrt[4]{\frac{15 \bullet 8 + 5}{8}} \bullet \sqrt[4]{\frac{5}{2}} = \sqrt[4]{\frac{125}{8} \bullet \frac{5}{2}} =\]

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

\[4)\ \sqrt[3]{11\frac{1}{4}}\ :\sqrt[3]{3\frac{1}{3}} =\]

\[= \sqrt[3]{\frac{11 \bullet 4 + 1}{4}}\ :\sqrt[3]{\frac{3 \bullet 3 + 1}{3}} =\]

\[= \sqrt[3]{\frac{45}{4} \bullet \frac{3}{10}} = \sqrt[3]{\frac{9 \bullet 3}{4 \bullet 2}} = \sqrt[3]{\frac{3^{2} \bullet 3}{2^{2} \bullet 2}} =\]

\[= \sqrt[3]{\frac{3^{3}}{2^{3}}} = \frac{3}{2} = 1,5\]

\[5)\ \left( \sqrt[3]{\sqrt{27}} \right)^{2} = \left( \sqrt[{3 \bullet 2}]{27} \right)^{2} =\]

\[= \sqrt[{3 \bullet 2}]{27^{2}} = \sqrt[3]{27} = \sqrt[3]{3^{3}} = 3\]

\[6)\ \left( \sqrt{\sqrt[3]{16}} \right)^{3} = \left( \sqrt[{2 \bullet 3}]{16} \right)^{3} =\]

\[= \sqrt[{2 \bullet 3}]{16^{3}} = \sqrt{16} = \sqrt{4^{2}} = 4\]