Страница 37 Проверь себя, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
1) \(\frac{15^{\frac{2}{3}} \cdot 3^{\frac{7}{3}}}{5^{-\frac{1}{3}}}\); 2) \(\left(\frac{4}{5}\right)^{-2} - \left(\frac{1}{27}\right)^{\frac{1}{3}} + 4 \cdot 379^0\); 3) \(\left(\sqrt[3]{128} + \sqrt[3]{\frac{1}{4}}\right) : \sqrt[3]{2}\).
1) \(\frac{15^{\frac{2}{3}} \cdot 3^{\frac{7}{3}}}{5^{-\frac{1}{3}}} = 5^{\frac{2}{3}+\frac{1}{3}} \cdot 3^{\frac{2}{3}+\frac{7}{3}} = 5 \cdot 3^{3} = 135\);
2) \(\left(\frac{4}{5}\right)^{-2} - \left(\frac{1}{27}\right)^{\frac{1}{3}} + 4 \cdot 379^{0} = \left(\frac{5}{4}\right)^{2} - \frac{1}{\sqrt[3]{3^{3}}} + 4 \cdot 1 = \frac{25}{16} - \frac{1}{3} + 4 =\)
\(= \dfrac{75 - 16}{48} + 4 = \dfrac{59}{48} + 4 = 1\dfrac{11}{48} + 4 = 5\dfrac{11}{48};\)
3) \(\left(\sqrt[3]{128} + \sqrt[3]{\frac{1}{4}}\right) : \sqrt[3]{2} = \sqrt[3]{128 : 2} + \sqrt[3]{\frac{1}{4} : 2} =\)
\(= \sqrt[3]{64} + \sqrt[3]{\frac{1}{8}} = 4 + \frac{1}{2} = 4{,}5\).
1) \(\sqrt[3]{\frac{ab^2}{c}} \cdot \sqrt[3]{\frac{a^5 b}{c^2}}\); 2) \(\frac{a^{-3} \cdot a^{\frac{7}{3}}}{a^{\frac{1}{3}}}\).
1) \(\sqrt[3]{\frac{ab^{2}}{c}} \cdot \sqrt[3]{\frac{a^{5}b}{c^{2}}} = \sqrt[3]{\frac{a^{6}b^{3}}{c^{3}}} = \frac{a^{2}b}{c}\);
2) \(\frac{a^{-3} \cdot a^{\frac{7}{3}}}{a^{\frac{1}{3}}} = a^{-3+\frac{7}{3}-\frac{1}{3}} = a^{-3+2} = a^{-1} = \frac{1}{a}\).
\(\dfrac{a - 9a^{\frac{1}{2}}}{7a^{\frac{1}{4}} + 21} = \dfrac{a^{\frac{1}{2}}\left(a^{\frac{1}{2}} - 9\right)}{7\left(a^{\frac{1}{4}} + 3\right)} = \dfrac{a^{\frac{1}{2}}\left(a^{\frac{1}{4}} - 3\right)\left(a^{\frac{1}{4}} + 3\right)}{7\left(a^{\frac{1}{4}} + 3\right)} = \dfrac{a^{\frac{1}{2}}\left(a^{\frac{1}{4}} - 3\right)}{7}.\)
\(\sqrt[5]{\left(\frac{2}{9}\right)^{3}} = \left(\frac{2}{9}\right)^{\frac{3}{5}}\), \(\sqrt[5]{\left(\frac{1}{4}\right)^{3}} = \left(\frac{1}{4}\right)^{\frac{3}{5}}\),
т.к. \(\frac{2}{9} < \frac{1}{4} = \frac{2}{8}\) и \(\frac{3}{5} > 0\), то \(\sqrt[5]{\left(\frac{2}{9}\right)^{3}} < \sqrt[5]{\left(\frac{1}{4}\right)^{3}}\).
\(\left(\sqrt[3]{a} + \sqrt[3]{b}\right)^{2} - \left(\sqrt[3]{a} - \sqrt[3]{b}\right)^{2} =\)
\(= a^{\frac{2}{3}} + 2a^{\frac{1}{3}}b^{\frac{1}{3}} + b^{\frac{2}{3}} - \left(a^{\frac{2}{3}} - 2a^{\frac{1}{3}}b^{\frac{1}{3}} + b^{\frac{2}{3}}\right) =\)
\(= a^{\frac{2}{3}} + 2a^{\frac{1}{3}}b^{\frac{1}{3}} + b^{\frac{2}{3}} - a^{\frac{2}{3}} + 2a^{\frac{1}{3}}b^{\frac{1}{3}} - b^{\frac{2}{3}} = 4a^{\frac{1}{3}}b^{\frac{1}{3}}.\)
