Страница 69 номер 177, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Расположить числа в порядке возрастания: 1) \(0{,}3^{\pi}\), \(0{,}3^{0{,}5}\), \(0{,}3^{\frac{2}{3}}\), \(0{,}3^{3{,}1415}\); 2) \((\sqrt{2})^{\pi}\), \(1{,}9^{\pi}\), \(\left(\frac{1}{\sqrt{2}}\right)^{\pi}\), \(\pi^{\pi}\); 3) \(5^{-2}\), \(5^{-0{,}7}\), \(5^{\frac{1}{3}}\), \(\left(\frac{1}{5}\right)^{2{,}1}\); 4) \(0{,}5^{-\frac{2}{3}}\), \(1{,}3^{-\frac{2}{3}}\), \(\pi^{-\frac{2}{3}}\), \((\sqrt{2})^{-\frac{2}{3}}\).
1) \(0{,}3^{\pi}\), \(0{,}3^{0{,}5}\), \(0{,}3^{\frac{2}{3}}\), \(0{,}3^{3{,}1415}\): \(0{,}3 < 1\) и \(\pi > 3{,}1415 > \frac{2}{3} > 0{,}5\). Тогда
\(0{,}3^{\pi} < 0{,}3^{3{,}1415} < 0{,}3^{\frac{2}{3}} < 0{,}3^{0{,}5};\)
2) \(\left( \sqrt{2} \right)^{\pi}\), \(1{,}9^{\pi}\), \(\left( \frac{1}{\sqrt{2}} \right)^{\pi}\), \(\pi^{\pi}\): \(\pi > 1{,}9 > \sqrt{2} > \frac{1}{\sqrt{2}}\) и \(\pi > 0\). Тогда
\(\left( \dfrac{1}{\sqrt{2}} \right)^{\pi} < \left( \sqrt{2} \right)^{\pi} < 1{,}9^{\pi} < \pi^{\pi};\)
3) \(5^{-2}\), \(5^{-0{,}7}\), \(5^{\frac{1}{3}}\), \(\left( \frac{1}{5} \right)^{2{,}1}\): \(\frac{1}{3} > -0{,}7 > -2 > -2{,}1\) и \(5 > 1\). Тогда
\(\left( \dfrac{1}{5} \right)^{2{,}1} = 5^{-2{,}1} < 5^{-2} < 5^{-0{,}7} < 5^{\frac{1}{3}};\)
4) \(0{,}5^{-\frac{2}{3}}\), \(1{,}3^{-\frac{2}{3}}\), \(\pi^{-\frac{2}{3}}\), \(\left( \sqrt{2} \right)^{-\frac{2}{3}}\): \(-\frac{2}{3} < 0\) и \(\pi > \sqrt{2} > 1{,}3 > 0{,}5\). Тогда
\(\pi^{-\frac{2}{3}} < \left( \sqrt{2} \right)^{-\frac{2}{3}} < 1{,}3^{-\frac{2}{3}} < 0{,}5^{-\frac{2}{3}}.\)
