Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 62

\[\boxed{\mathbf{62}\mathbf{.}}\]

\[1)\ a^{\frac{1}{3}} \bullet \sqrt{a} = a^{\frac{1}{3}} \bullet a^{\frac{1}{2}} = a^{\frac{1}{3} + \frac{1}{2}} =\]

\[= a^{\frac{2 + 3}{6}} = a^{\frac{5}{6}}\]

\[2)\ b^{\frac{1}{2}} \bullet b^{\frac{1}{3}} \bullet \sqrt[6]{b} = b^{\frac{1}{2}} \bullet b^{\frac{1}{3}} \bullet b^{\frac{1}{6}} =\]

\[= b^{\frac{1}{2} + \frac{1}{3} + \frac{1}{6}} = b^{\frac{3 + 2 + 1}{6}} = b^{\frac{6}{6}} =\]

\[= b^{1} = b\]

\[3)\ \sqrt[3]{b}\ :b^{\frac{1}{6}} = b^{\frac{1}{3}}\ :b^{\frac{1}{6}} = b^{\frac{1}{3} - \frac{1}{6}} =\]

\[= b^{\frac{2 - 1}{6}} = b^{\frac{1}{6}}\]

\[4)\ a^{\frac{4}{3}}\ :\ \sqrt[3]{a} = a^{\frac{4}{3}}\ :a^{\frac{1}{3}} = a^{\frac{4}{3} - \frac{1}{3}} =\]

\[= a^{\frac{3}{3}} = a^{1} = a\]

\[5)\ x^{1,7} \bullet x^{2,8}\ :\sqrt{x^{5}} =\]

\[= x^{1,7} \bullet x^{2,8}\ :x^{\frac{5}{2}} =\]

\[= x^{1,7} \bullet x^{2,8}\ :x^{2,5} =\]

\[= x^{1,7 + 2,8 - 2,5} = x^{2}\]

\[6)\ y^{- 3,8}\ :y^{- 2,3} \bullet \sqrt[3]{y} =\]

\[= y^{- \frac{38}{20}}\ :y^{- \frac{23}{10}} \bullet y^{\frac{1}{3}} =\]

\[= y^{- \frac{38}{10} - \left( - \frac{23}{10} \right) + \frac{1}{3}} = y^{\frac{23}{10} - \frac{38}{10} + \frac{1}{3}} =\]

\[= y^{- \frac{15}{10} + \frac{1}{3}} = y^{- \frac{3}{2} + \frac{1}{3}} = y^{\frac{- 9 + 2}{6}} =\]

\[= y^{- \frac{7}{6}} = y^{- 1\frac{1}{6}}\]