ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 246

\[{\boxed{\mathbf{246}\mathbf{.}} }{1)\ 4^{- \sqrt{3}}\text{\ \ }и\ \ 4^{- \sqrt{2}}}\]

\[\sqrt{3} > \sqrt{2}\]

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

\[4^{- \sqrt{3}} < 4^{- \sqrt{2}}.\]

\[2)\ 2^{\sqrt{3}}\text{\ \ }и\ \ 2^{1,7}\]

\[300 > 289\]

\[\sqrt{300} > 17\]

\[\sqrt{3} > 1,7\]

\[2^{\sqrt{3}} > 2^{1,7}.\]

\[3)\ \left( \frac{1}{2} \right)^{1,4}\text{\ \ }и\ \ \left( \frac{1}{2} \right)^{\sqrt{2}}\]

\[196 < 200\]

\[14 < \sqrt{200}\]

\[1,4 < \sqrt{2}\]

\[\left( \frac{1}{2} \right)^{1,4} > \left( \frac{1}{2} \right)^{\sqrt{2}}.\]

\[4)\ \left( \frac{1}{9} \right)^{\pi}\text{\ \ }и\ \ \left( \frac{1}{9} \right)^{3,14}\]

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

\[\pi > 3,14\]

\[\left( \frac{1}{9} \right)^{\pi} < \left( \frac{1}{9} \right)^{3,14}.\]