ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 455

\[\boxed{\mathbf{455}.}\]

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

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

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

\[= \frac{7 \bullet 2}{5} = \frac{14}{5} = 2\frac{4}{5} = 2,8;\]

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

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

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

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

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

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

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

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

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

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

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

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

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

\[+ 3 - 4 = 1,5 - 1 = 0,5;\]

\[5)\ \sqrt[4]{17 - \sqrt{33}} \bullet \sqrt[4]{17 + \sqrt{33}} =\]

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

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

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

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