Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 76

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\[= 2^{3} + 3 \bullet 10 - \left( \frac{3^{5}}{2^{5}} \right)^{\frac{1}{5}} =\]

\[= 8 + 30 - \frac{3}{2} = 38 - 1,5 = 36,5\]

\[2)\ 27^{\frac{2}{3}} - ( - 2)^{- 2} + \left( 3\frac{3}{8} \right)^{- \frac{1}{3}} =\]

\[= 3^{2} - \frac{1}{4} + \left( \frac{27}{8} \right)^{- \frac{1}{3}} =\]

\[= \frac{36}{4} - \frac{1}{4} + \left( \frac{8}{27} \right)^{\frac{1}{3}} =\]

\[= \frac{35}{4} + \left( \frac{2^{3}}{3^{3}} \right)^{\frac{1}{3}} = \frac{35}{4} + \frac{2}{3} =\]

\[= \frac{105 + 8}{12} = \frac{113}{12} = 9\frac{5}{12}\]

\[3)\ (0,001)^{- \frac{1}{3}} - 2^{- 2} \bullet 64^{\frac{2}{3}} - 8^{- 1\frac{1}{3}} =\]

\[= \left( (0,1)^{3} \right)^{- \frac{1}{3}} - \left( \frac{1}{2} \right)^{2} \bullet \left( 4^{3} \right)^{\frac{2}{3}} - 8^{- \frac{4}{3}} =\]

\[= (0,1)^{- 1} - \frac{1}{4} \bullet 4^{2} - \left( 2^{3} \right)^{- \frac{4}{3}} =\]

\[= \left( \frac{1}{10} \right)^{- 1} - 4^{2 - 1} - 2^{- 4} =\]

\[= 10 - 4 - \left( \frac{1}{2} \right)^{4} =\]

\[= 6 - \frac{1}{16} = 5\frac{15}{16}\]

\[= ( - 2)^{4} - 5^{1} - \left( \frac{9}{4} \right)^{- \frac{3}{2}} =\]

\[= 16 - 5 - \left( \frac{4}{9} \right)^{\frac{3}{2}} =\]

\[= 11 - \left( \frac{2^{2}}{3^{2}} \right)^{\frac{3}{2}} = 11 - \frac{2^{3}}{3^{3}} =\]

\[= 11 - \frac{8}{27} = 10\frac{19}{27}\]