ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 47

\[\boxed{\mathbf{47}\mathbf{.}}\]

\[1)\ \frac{\sqrt[3]{49} \bullet \sqrt[3]{112}}{\sqrt[3]{250}} = \sqrt[3]{\frac{49 \bullet 112}{250}} =\]

\[= \sqrt[3]{\frac{7 \bullet 7 \bullet 16 \bullet 7}{125 \bullet 2}} = \sqrt[3]{\frac{7^{3} \bullet 2^{4}}{5^{3} \bullet 2}} =\]

\[= \sqrt[3]{\frac{7^{3} \bullet 2^{3}}{5^{3}}} = \frac{7 \bullet 2}{5} = \frac{14}{5} =\]

\[= 2\frac{4}{5} = 2,8\]

\[2)\ \frac{\sqrt[4]{54} \bullet \sqrt[4]{120}}{\sqrt[4]{5}} = \sqrt[4]{\frac{54 \bullet 120}{5}} =\]

\[= \sqrt[4]{54 \bullet 24} = \sqrt[4]{6 \bullet 9 \bullet 6 \bullet 4} =\]

\[= \sqrt[4]{6^{2} \bullet 36} = \sqrt[4]{6^{2} \bullet 6^{2}} = \sqrt[4]{6^{4}} =\]

\[= 6\]

\[3)\frac{\sqrt[4]{32}}{\sqrt[4]{2}} + \sqrt[6]{27^{2}} - \sqrt{\sqrt[3]{64}} =\]

\[= \sqrt[4]{\frac{32}{2}} + \sqrt[6]{\left( 3^{3} \right)^{2}} - \sqrt[{2 \bullet 3}]{64} =\]

\[= \sqrt[4]{16} + \sqrt[6]{3^{6}} - \sqrt[6]{2^{6}} =\]

\[= \sqrt[4]{2^{4}} + 3 - 2 = 2 + 1 = 3\]

\[4)\ \sqrt[3]{3\frac{3}{8}} + \sqrt[4]{18} \bullet \sqrt[4]{4\frac{1}{2}} - \sqrt{\sqrt{256}} =\]

\[= \sqrt[3]{\frac{27}{8}} + \sqrt[4]{9 \bullet 9} - \sqrt[4]{256} =\]

\[= \sqrt[3]{\frac{3^{3}}{2^{3}}} + \sqrt[4]{3^{4}} - \sqrt[4]{4^{4}} =\]

\[= \frac{3}{2} + 3 - 4 = 1,5 - 1 = 0,5\]

\[5)\ \sqrt[3]{11 - \sqrt{57}} \bullet \sqrt[3]{11 + \sqrt{57}} =\]

\[= \sqrt[3]{\left( 11 - \sqrt{57} \right)\left( 11 + \sqrt{57} \right)} =\]

\[= \sqrt[3]{11^{2} - \left( \sqrt{57} \right)^{2}} =\]

\[= \sqrt[3]{121 - 57} = \sqrt[3]{64} = \sqrt[3]{4^{3}} = 4\]

\[6)\ \sqrt[4]{17 - \sqrt{33}} \bullet \sqrt[4]{17 + \sqrt{33}} =\]

\[= \sqrt[4]{\left( 17 - \sqrt{33} \right)\left( 17 + \sqrt{33} \right)} =\]

\[= \sqrt[4]{17^{2} - \left( \sqrt{33} \right)^{2}} =\]

\[= \sqrt[4]{289 - 33} = \sqrt[4]{256} =\]

\[= \sqrt[4]{4^{4}} = 4\]