ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 727

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\[1)\ 2^{- \sqrt{5}}:\]

\[- \sqrt{5} < 0\]

\[2^{- \sqrt{5}} < 2^{0}\]

\[2^{- \sqrt{5}} < 1.\]

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

\[\sqrt{3} > 0\]

\[\left( \frac{1}{2} \right)^{\sqrt{3}} < \left( \frac{1}{2} \right)^{0}\]

\[\left( \frac{1}{2} \right)^{\sqrt{3}} < 1.\]

\[3)\ \left( \frac{\pi}{4} \right)^{\sqrt{5} - 2}:\]

\[\pi \approx 3,14\ldots,\ значит\ \frac{\pi}{4} < 1\ \]

\[5 > 4\]

\[\sqrt{5} > 2\]

\[\sqrt{5} - 2 > 0\]

\[\left( \frac{\pi}{4} \right)^{\sqrt{5} - 2} < \left( \frac{\pi}{4} \right)^{0}\]

\[\left( \frac{\pi}{4} \right)^{\sqrt{5} - 2} < 1.\]

\[4)\ \left( \frac{1}{3} \right)^{\sqrt{8} - 3}:\]

\[8 < 9\]

\[\sqrt{8} < 3\]

\[\sqrt{8} - 3 < 0\]

\[\left( \frac{1}{3} \right)^{\sqrt{8} - 3} > \left( \frac{1}{3} \right)^{0}\]

\[\left( \frac{1}{3} \right)^{\sqrt{8} - 3} > 1.\]