Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 124

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\[1)\ {3,1}^{7}\text{\ \ }и\ \ {4,3}^{7}\]

\[3,1 < 4,3\]

\[{3,1}^{7} < {4,3}^{7}.\]

\[2)\ \left( \frac{10}{11} \right)^{3}\text{\ \ }и\ \ \left( \frac{12}{11} \right)^{3}\]

\[\frac{10}{11} < \frac{12}{11}\]

\[\left( \frac{10}{11} \right)^{3} < \left( \frac{12}{11} \right)^{3}.\]

\[3)\ {0,3}^{8}\text{\ \ }и\ \ {0,2}^{8}\]

\[0,3 > 0,2\]

\[{0,3}^{8} > {0,2}^{8}.\]

\[4)\ {2,5}^{2}\text{\ \ }и\ \ {2,6}^{2}\]

\[2,5 < 2,6\]

\[{2,5}^{2} < {2,6}^{2}.\]

\[5)\ \left( \frac{7}{9} \right)^{- 2}\text{\ \ }и\ \ \left( \frac{8}{10} \right)^{- 2}\]

\[\frac{7}{9} = \frac{70}{90}\text{\ \ }и\ \ \frac{8}{10} = \frac{72}{90}\]

\[\frac{70}{90} < \frac{72}{90}\]

\[\frac{7}{9} < \frac{8}{10}\]

\[\left( \frac{7}{9} \right)^{- 2} > \left( \frac{8}{10} \right)^{- 2}.\]

\[6)\ \left( \frac{14}{15} \right)^{- 6}\text{\ \ }и\ \ \left( \frac{15}{16} \right)^{- 6}\]

\[\frac{14}{15} = \frac{224}{240}\text{\ \ }и\ \ \frac{15}{16} = \frac{225}{240}\]

\[\frac{224}{240} < \frac{225}{240}\]

\[\frac{14}{15} < \frac{15}{16}\]

\[\left( \frac{14}{15} \right)^{- 6} > \left( \frac{15}{16} \right)^{- 6}.\]

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

\[\left( 4\sqrt{3} \right)^{2} = 16 \bullet 3 = 48\ \ и\ \ \]

\[\left( 3\sqrt{4} \right)^{2} = 9 \bullet 4 = 36\]

\[48 > 36\]

\[4\sqrt{3} > 3\sqrt{4}\]

\[\left( 4\sqrt{3} \right)^{- 3} < \left( 3\sqrt{4} \right)^{- 3}.\]

\[8)\ \left( 2\sqrt[3]{6} \right)^{- 5}\text{\ \ }и\ \ \left( 6\sqrt[3]{2} \right)^{- 5}\]

\[\left( 2\sqrt[3]{6} \right)^{3} = 8 \bullet 6 = 48\ \ и\ \ \]

\[\left( 6\sqrt[3]{2} \right)^{3} = 216 \bullet 2 = 432\]

\[48 < 432\]

\[2\sqrt[3]{6} < 6\sqrt[3]{2}\]

\[\left( 2\sqrt[3]{6} \right)^{- 5} > \left( 6\sqrt[3]{2} \right)^{- 5}.\]