Страница 33 номер 78, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Упростить выражение (77–78). 1) \(\frac{a^{\frac{4}{3}}\left(a^{-\frac{1}{3}} + a^{\frac{2}{3}}\right)}{a^{\frac{1}{4}}\left(a^{\frac{3}{4}} + a^{-\frac{1}{4}}\right)}\); 2) \(\frac{b^{\frac{1}{5}}\left(\sqrt[5]{b^4} - \sqrt[5]{b^{-1}}\right)}{b^{\frac{2}{3}}\left(\sqrt[3]{b} - \sqrt[3]{b^{-2}}\right)}\); 3) \(\frac{a^{\frac{5}{3}}b^{-1} - a^{-\frac{1}{3}}}{\sqrt[3]{a^2}}\); 4) \(\frac{a^{\frac{1}{3}}\sqrt{b} + b^{\frac{1}{3}}\sqrt{a}}{\sqrt[6]{a} + \sqrt[6]{b}}\).
1) \(\frac{a^{\frac{4}{3}}\left(a^{-\frac{1}{3}} + a^{\frac{2}{3}}\right)}{a^{\frac{1}{4}}\left(a^{\frac{3}{4}} + a^{-\frac{1}{4}}\right)} = \frac{a^{\frac{4}{3} - \frac{1}{3}}(1 + a)}{a^{\frac{1}{4} - \frac{1}{4}}(a + 1)} = \frac{a^{1}}{a^{0}} = a;\)
2) \(\frac{b^{\frac{1}{5}}\left(\sqrt[5]{b^{4}} - \sqrt[5]{b^{-1}}\right)}{b^{\frac{2}{3}}\left(\sqrt[3]{b} - \sqrt[3]{b^{-2}}\right)} = \frac{b^{\frac{1}{5}}\left(b^{\frac{4}{5}} - b^{-\frac{1}{5}}\right)}{b^{\frac{2}{3}}\left(b^{\frac{1}{3}} - b^{-\frac{2}{3}}\right)} = \frac{b^{\frac{1}{5} - \frac{1}{5}}(b - 1)}{b^{\frac{2}{3} - \frac{2}{3}}(b - 1)} = \frac{b^{0}}{b^{0}} = 1;\)
3) \(\frac{a^{\frac{5}{3}}b^{-1} - a^{-\frac{1}{3}}}{\sqrt[3]{a^{2}}} = \frac{a^{\frac{2}{3}}(ab^{-1} - a^{-1})}{a^{\frac{2}{3}}} = ab^{-1} - a^{-1};\)
4) \(\frac{a^{\frac{1}{3}}\sqrt{b} + b^{\frac{1}{3}}\sqrt{a}}{\sqrt[6]{a} + \sqrt[6]{b}} = \frac{a^{\frac{1}{3}}b^{\frac{1}{3}}\left(b^{\frac{1}{6}} + a^{\frac{1}{6}}\right)}{a^{\frac{1}{6}} + b^{\frac{1}{6}}} = a^{\frac{1}{3}}b^{\frac{1}{3}}.\)
