Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 99

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\[1)\ {0,88}^{\frac{1}{6}}\text{\ \ }и\ \ \left( \frac{6}{11} \right)^{\frac{1}{6}};\]

\[0,88 = \frac{88}{100} = \frac{22}{25} = \frac{242}{275}\text{\ \ }и\ \ \]

\[\frac{6}{11} = \frac{150}{275};\]

\[\frac{242}{275} > \frac{150}{275};\]

\[0,88 > \frac{6}{11};\]

\[{0,88}^{\frac{1}{6}} > \left( \frac{6}{11} \right)^{\frac{1}{6}}.\]

\[2)\ \left( \frac{5}{12} \right)^{- \frac{1}{4}}\ и\ \ {0,41}^{- \frac{1}{4}};\]

\[\frac{5}{12} = \frac{500}{1200}\ \ и\ \ \]

\[0,41 = \frac{41}{100} = \frac{492}{1200};\]

\[\frac{500}{1200} > \frac{492}{1200};\]

\[\frac{5}{12} > 0,41;\]

\[\left( \frac{5}{12} \right)^{- \frac{1}{4}} < (0,41)^{- \frac{1}{4}}.\]

\[3)\ {4,09}^{\sqrt[3]{2}}\text{\ \ }и\ \ \left( 4\frac{3}{25} \right)^{\sqrt[3]{2}};\]

\[4,09 = \frac{409}{100}\text{\ \ }и\ \ 4\frac{3}{25} =\]

\[= \frac{4 \bullet 25 + 3}{25} = \frac{103}{25} = \frac{412}{100};\]

\[\frac{409}{100} < \frac{412}{100};\]

\[4,09 < 4\frac{3}{25};\]

\[\ {4,09}^{\sqrt[3]{2}} < \left( 4\frac{3}{25} \right)^{\sqrt[3]{2}}.\]

\[4)\ \left( \frac{11}{12} \right)^{- \sqrt{5}}\ и\ \ \left( \frac{12}{13} \right)^{- \sqrt{5}};\]

\[\frac{11}{12} = \frac{143}{156}\text{\ \ }и\ \ \frac{12}{13} = \frac{144}{156};\]

\[\frac{143}{156} < \frac{144}{156};\]

\[\frac{11}{12} < \frac{12}{13};\]

\[\left( \frac{11}{12} \right)^{- \sqrt{5}} > \left( \frac{12}{13} \right)^{- \sqrt{5}}.\]

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