ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 102

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\[1)\ \sqrt[7]{\left( \frac{1}{2} - \frac{1}{3} \right)^{2}}\text{\ \ }и\ \ \sqrt[7]{\left( \frac{1}{3} - \frac{1}{4} \right)^{2}};\]

\[\sqrt[7]{\left( \frac{1}{2} - \frac{1}{3} \right)^{2}} = \left( \frac{1}{2} - \frac{1}{3} \right)^{\frac{2}{7}} =\]

\[= \left( \frac{3 - 2}{6} \right)^{\frac{2}{7}} = \left( \frac{1}{6} \right)^{\frac{2}{7}} = \left( 6^{- 1} \right)^{\frac{2}{7}} =\]

\[= 6^{- \frac{2}{7}};\]

\[\sqrt[7]{\left( \frac{1}{3} - \frac{1}{4} \right)^{2}} = \left( \frac{1}{3} - \frac{1}{4} \right)^{\frac{2}{7}} =\]

\[= \left( \frac{4 - 3}{12} \right)^{\frac{2}{7}} = \left( \frac{1}{12} \right)^{\frac{2}{7}} = \left( 12^{- 1} \right)^{\frac{2}{7}} =\]

\[= 12^{- \frac{2}{7}};\]

\[6 < 12;\]

\[6^{- \frac{2}{7}} > 12^{- \frac{2}{7}};\]

\[\ \sqrt[7]{\left( \frac{1}{2} - \frac{1}{3} \right)^{2}} > \sqrt[7]{\left( \frac{1}{3} - \frac{1}{4} \right)^{2}}.\]

\[2)\ \sqrt[7]{\left( 1\frac{1}{4} - 1\frac{1}{5} \right)^{2}}\text{\ \ }и\ \]

\[\ \sqrt[7]{\left( 1\frac{1}{6} - 1\frac{1}{7} \right)^{2}};\]

\[\ \sqrt[7]{\left( 1\frac{1}{4} - 1\frac{1}{5} \right)^{2}} = \left( 1\frac{1}{4} - 1\frac{1}{5} \right)^{\frac{2}{7}} =\]

\[= \left( \frac{1}{4} - \frac{1}{5} \right)^{\frac{2}{7}} = \left( \frac{5 - 4}{20} \right)^{\frac{2}{7}} =\]

\[= \left( \frac{1}{20} \right)^{\frac{2}{7}} = 20^{- \frac{2}{7}};\]

\[\sqrt[7]{\left( 1\frac{1}{6} - 1\frac{1}{7} \right)^{2}} = \left( 1\frac{1}{6} - 1\frac{1}{7} \right)^{\frac{2}{7}} =\]

\[= \left( \frac{1}{6} - \frac{1}{7} \right)^{\frac{2}{7}} = \left( \frac{7 - 6}{42} \right)^{\frac{2}{7}} =\]

\[= \left( \frac{1}{42} \right)^{\frac{2}{7}} = 42^{- \frac{2}{7}};\]

\[20 < 42;\]

\[20^{- \frac{2}{7}} > 42^{- \frac{2}{7}};\]

\[\text{\ \ }\sqrt[7]{\left( 1\frac{1}{4} - 1\frac{1}{5} \right)^{2}} > \sqrt[7]{\left( 1\frac{1}{6} - 1\frac{1}{7} \right)^{2}}\text{\ .}\]