ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 644

\[\boxed{\mathbf{644}.}\]

\[Если\ основание < 1,\ то\ больше\ та\ функция,\ у\ которой\ степень\ меньше.\]

\[1)\ {0,3}^{\pi},\ \ \ {0,3}^{0,5},\ \ \ {0,3}^{\frac{2}{3}},\ \ \ {0,3}^{3,1315};\]

\[\pi \approx 3,1415926\ldots;\]

\[0,5 = \frac{1}{2} = \frac{3}{6};\]

\[\frac{2}{3} = \frac{4}{6};\]

\[0,5 < \frac{2}{3} < 3,1415 < \pi;\]

\[Ответ:\ \ {0,3}^{\pi},\ \ \ {0,3}^{3,1415},\ \ \ {0,3}^{\frac{2}{3}},\ \ \ \]

\[{0,3}^{0,5}.\]

\[2)\ \sqrt{2^{\pi}},\ \ \ {1,9}^{\pi},\ \ \ \left( \frac{1}{\sqrt{2}} \right)^{\pi},\ \ \ \pi^{\pi};\]

\[196 < 200 < 225\ \ \ = > \ \ \ 14 <\]

\[< \sqrt{200} < 15\ \ \ = > \ \ \ 1,4 <\]

\[< \sqrt{2} < 1,5;\]

\[\sqrt{2} > 1\ \ \ = > \ \ \ \frac{1}{\sqrt{2}} < 1;\]

\[\pi \approx 3,1415\ldots;\]

\[\frac{1}{\sqrt{2}} < \sqrt{2} < 1,9 < \pi;\]

\[Ответ:\ \ \left( \frac{1}{\sqrt{2}} \right)^{\pi},\ \ \ \sqrt{2^{\pi}},\ \ \ {1,9}^{\pi},\ \ \ \pi^{\pi}.\]

\[3)\ 5^{- 2},\ \ \ 5^{- 0,7},\ \ \ 5^{\frac{1}{3}},\ \ \ \left( \frac{1}{5} \right)^{2,1};\]

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

\[- 2,1 < - 2 < - 0,7 < \frac{1}{3};\]

\[Ответ:\ \ \left( \frac{1}{5} \right)^{2,1},\ \ \ 5^{- 2},\ \ \ 5^{- 0,7},\ \ \ 5^{\frac{1}{3}}.\]

\[4)\ {0,5}^{- \frac{2}{3}},\ \ \ {1,3}^{- \frac{2}{3}},\ \ \ \pi^{- \frac{2}{3}},\ \ \ \left( \sqrt{2} \right)^{- \frac{2}{3}};\]

\[196 < 200 < 225\ \ \ = > \ \ \ 14 <\]

\[< \sqrt{200} < 15\ \ \ = > \ \ \ 1,4 <\]

\[< \sqrt{2} < 1,5;\]

\[\pi \approx 3,1415\ldots;\]

\[0,5 < 1,3 < \sqrt{2} < \pi;\]

\[Ответ:\ \ \pi^{- \frac{2}{3}},\ \ \ \left( \sqrt{2} \right)^{- \frac{2}{3}},\ \ \ {1,3}^{- \frac{2}{3}},\ \ \ \]

\[{0,5}^{- \frac{2}{3}}.\]