Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1250

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\[1)\ {2,5}^{\frac{1}{7}}\text{\ \ }и\ \ {2,5}^{0,5}\]

\[\frac{{2,5}^{\frac{1}{7}}}{{2,5}^{0,5}} = \frac{{2,5}^{\frac{1}{7}}}{{2,5}^{\frac{1}{2}}} = {2,5}^{\frac{1}{7} - \frac{1}{2}} =\]

\[= {2,5}^{\frac{2}{14} - \frac{7}{14}} = {2,5}^{- \frac{5}{14}} =\]

\[= \left( \frac{5}{2} \right)^{- \frac{5}{14}} = \left( \frac{2}{5} \right)^{\frac{5}{14}} < 1;\]

\[{2,5}^{\frac{1}{7}} < {2,5}^{0,5}.\]

\[2)\ {0,2}^{\frac{2}{3}}\text{\ \ }и\ \ {0,2}^{\frac{3}{4}}\]

\[\frac{{0,2}^{\frac{2}{3}}}{{0,2}^{\frac{3}{4}}} = {0,2}^{\frac{2}{3} - \frac{3}{4}} = {0,2}^{\frac{8}{12} - \frac{9}{12}} =\]

\[= {0,2}^{- \frac{1}{12}} = \left( \frac{1}{5} \right)^{- \frac{1}{12}} = 5^{\frac{1}{12}} > 1;\]

\[{0,2}^{\frac{2}{3}} > {0,2}^{\frac{3}{4}}.\]

\[3)\ \log_{3,1}\sqrt{10}\text{\ \ }и\ \ \log_{3,1}3\]

\[\log_{3,1}x - возрастающая:\]

\[\frac{\sqrt{10}}{3} = \frac{\sqrt{10}}{\sqrt{9}} = \sqrt{\frac{10}{9}} > 1;\]

\[\log_{3,1}\sqrt{10} > \log_{3,1}3.\]

\[4)\ \log_{0,3}\frac{4}{5}\text{\ \ }и\ \ \log_{0,3}\frac{3}{4}\text{\ \ }\]

\[\log_{0,3}\frac{4}{5} - убывающая:\]

\[\frac{4}{5}\ :\frac{3}{4} = \frac{4}{5} \bullet \frac{4}{3} = \frac{16}{15} > 1;\]

\[\log_{0,3}\frac{4}{5} < \log_{0,3}\frac{3}{4}.\]