ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 27

\[\boxed{\mathbf{27}\mathbf{.}}\]

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

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

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

\[\sqrt{0,81} = \sqrt{\frac{81}{100}} = \sqrt{\frac{9 \bullet 9}{10 \bullet 10}} =\]

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

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

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

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

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

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

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

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

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

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

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

\[3)\ \sqrt[4]{0} = \sqrt[4]{0 \bullet 0 \bullet 0 \bullet 0} = 0;\]

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

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

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

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

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

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