Страница 36 номер 99, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Сравнить числа: 1) \(0{,}88^{\frac{1}{6}}\) и \(\left(\frac{6}{11}\right)^{\frac{1}{6}}\); 2) \(\left(\frac{5}{12}\right)^{-\frac{1}{4}}\) и \(0{,}41^{-\frac{1}{4}}\); 3) \(4{,}09^{\sqrt[3]{2}}\) и \(\left(4\frac{3}{25}\right)^{\sqrt[3]{2}}\); 4) \(\left(\frac{11}{12}\right)^{-\sqrt{5}}\) и \(\left(\frac{12}{13}\right)^{-\sqrt{5}}\).
1) \(0{,}88^{\frac{1}{6}} > \left(\frac{6}{11}\right)^{\frac{1}{6}}\), т.к. \(\frac{1}{6} > 0\) и \(0{,}88 > \frac{6}{11}\);
2) \(\left(\frac{5}{12}\right)^{-\frac{1}{4}} < 0{,}41^{-\frac{1}{4}}\), т.к. \(\frac{5}{12} = \frac{40}{96} > \frac{41}{100}\) и \(-\frac{1}{4} < 0\);
3) \(4{,}09^{\sqrt[3]{2}} < \left(4\frac{3}{25}\right)^{\sqrt[3]{2}}\), т.к. \(4\frac{3}{25} = 4{,}12 > 4{,}09\) и \(\sqrt[3]{2} > 0\);
4) \(\left(\frac{11}{12}\right)^{-\sqrt{5}} = \left(1\frac{1}{11}\right)^{\sqrt{5}} > \left(\frac{12}{13}\right)^{-\sqrt{5}} = \left(1\frac{1}{12}\right)^{\sqrt{5}}\), т.к. \(1\frac{1}{11} > 1\frac{1}{12}\) и \(\sqrt{5} > 0\).
