ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 435

\[\boxed{\mathbf{435}.}\]

\[1)\ 1;\ 0;16;\ 0,81;\ 169;\ \frac{1}{289};\]

\[\sqrt{1} = \sqrt{1 \bullet 1} = 1;\]

\[\sqrt{0} = \sqrt{0 \bullet 0} = 0;\]

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

\[\sqrt{0,81} = \sqrt{\frac{81}{100}} =\]

\[= \sqrt{\frac{9 \bullet 9}{10 \bullet 10}} = \frac{9}{10} = 0,9;\]

\[\sqrt{169} = \sqrt{13 \bullet 13} = 13;\]

\[\sqrt{\frac{1}{289}} = \sqrt{\frac{1}{17} \bullet \frac{1}{17}} = \frac{1}{17}.\]

\[2)\ 1;\ 0;\ 125;\ \frac{1}{27};\ 0,027;\ 0,064;\]

\[\sqrt[3]{1} = \sqrt[3]{1 \bullet 1 \bullet 1} = 1;\]

\[\sqrt[3]{0} = \sqrt[3]{0 \bullet 0 \bullet 0} = 0;\]

\[\sqrt[3]{125} = \sqrt[3]{5 \bullet 5 \bullet 5} = 5;\]

\[\sqrt[3]{\frac{1}{27}} = \sqrt[3]{\frac{1}{3} \bullet \frac{1}{3} \bullet \frac{1}{3}} = \frac{1}{3};\]

\[\sqrt[3]{0,027} = \sqrt[3]{\frac{27}{1000}} =\]

\[= \sqrt[3]{\frac{3 \bullet 3 \bullet 3}{10 \bullet 10 \bullet 10}} = \frac{3}{10} = 0,3;\]

\[\sqrt[3]{0,064} = \sqrt[3]{\frac{64}{1000}} =\]

\[= \sqrt[3]{\frac{4 \bullet 4 \bullet 4}{10 \bullet 10 \bullet 10}} = \frac{4}{10} = 0,4.\]

\[3)\ 0;\ 1;\ 16;\ \frac{16}{81};\ \frac{256}{625};\ 0,0016;\]

\[\sqrt[4]{0} = \sqrt[4]{0 \bullet 0 \bullet 0 \bullet 0} = 0;\]

\[\sqrt[4]{1} = \sqrt[4]{1 \bullet 1 \bullet 1 \bullet 1} = 1;\]

\[\sqrt[4]{16} = \sqrt[4]{2 \bullet 2 \bullet 2 \bullet 2} = 2;\]

\[\sqrt[4]{\frac{16}{81}} = \sqrt[4]{\frac{2}{3} \bullet \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{2}{3}} = \frac{2}{3};\]

\[\sqrt[4]{\frac{256}{625}} = \sqrt[4]{\frac{4}{5} \bullet \frac{4}{5} \bullet \frac{4}{5} \bullet \frac{4}{5}} = \frac{4}{5};\]

\[\sqrt[4]{0,0016} = \sqrt[4]{\frac{16}{10\ 000}} =\]

\[= \sqrt[4]{\frac{2 \bullet 2 \bullet 2 \bullet 2}{10 \bullet 10 \bullet 10 \bullet 10}} = \frac{2}{10} = 0,2.\]

\[\boxed{\mathbf{436}.}\]

\[1)\ \sqrt[6]{36^{3}} = \sqrt[6]{\left( 6^{2} \right)^{3}} = \sqrt[6]{6^{6}} = 6;\]

\[2)\ \sqrt[12]{64^{2}} = \sqrt[12]{\left( 2^{6} \right)^{2}} =\]

\[= \sqrt[12]{2^{12}} = 2;\]

\[3)\ \sqrt[4]{\left( \frac{1}{25} \right)^{2}} = \sqrt[{2 \bullet 2}]{\left( \frac{1}{25} \right)^{2 \bullet 1}} =\]

\[= \sqrt{\frac{1}{25}} = \frac{1}{5};\]

\[4)\ \sqrt[8]{225^{4}} = \sqrt[8]{\left( 15^{2} \right)^{4}} =\]

\[= \sqrt[8]{15^{8}} = 15.\]