Страница 36 номер 102, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Сравнить числа: 1) \(\sqrt[7]{\left(\frac{1}{2} - \frac{1}{3}\right)^2}\) и \(\sqrt[7]{\left(\frac{1}{3} - \frac{1}{4}\right)^2}\); 2) \(\sqrt[5]{\left(1\frac{1}{4} - 1\frac{1}{5}\right)^3}\) и \(\sqrt[5]{\left(1\frac{1}{6} - 1\frac{1}{7}\right)^3}\).
1) \(\sqrt[7]{\left(\frac{1}{2} - \frac{1}{3}\right)^{2}} = \sqrt[7]{\left(\frac{1}{6}\right)^{2}} = \left(\frac{1}{6}\right)^{\frac{2}{7}}\), \(\sqrt[7]{\left(\frac{1}{3} - \frac{1}{4}\right)^{2}} = \sqrt[7]{\left(\frac{1}{12}\right)^{2}} = \left(\frac{1}{12}\right)^{\frac{2}{7}}\),
\(\frac{1}{6} > \frac{1}{12}\). Следовательно, \(\sqrt[7]{\left(\frac{1}{2} - \frac{1}{3}\right)^{2}} > \sqrt[7]{\left(\frac{1}{3} - \frac{1}{4}\right)^{2}}\).
2) \(\sqrt[5]{\left(1\frac{1}{4} - 1\frac{1}{5}\right)^{3}} = \sqrt[5]{\left(\frac{1}{4} - \frac{1}{5}\right)^{3}} = \left(\frac{1}{20}\right)^{\frac{3}{5}}\),
\(\sqrt[5]{\left(1\frac{1}{6} - 1\frac{1}{7}\right)^{3}} = \sqrt[5]{\left(\frac{1}{6} - \frac{1}{7}\right)^{3}} = \left(\frac{1}{42}\right)^{\frac{3}{5}}\), \(\frac{1}{20} > \frac{1}{42}\). Следовательно, \(\sqrt[5]{\left(1\frac{1}{4} - 1\frac{1}{5}\right)^{3}} > \sqrt[5]{\left(1\frac{1}{6} - 1\frac{1}{7}\right)^{3}}\)
