Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 96

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\[1)\ \frac{3}{4} - \left( \frac{2}{3} \right)^{- 1} = \frac{3}{4} - \frac{3}{2} = \frac{3}{4} - \frac{6}{4} =\]

\[= - \frac{3}{4} = - 0,75\]

\[2)\ \left( \frac{1}{27} \bullet 125^{- 1} \right)^{- \frac{1}{3}} =\]

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

\[= \left( 3^{- 3} \bullet 5^{- 3} \right)^{- \frac{1}{3}} = 3 \bullet 5 = 15\]

\[3)\ 27^{\frac{2}{3}} + 9^{- 1} = \left( 3^{3} \right)^{\frac{2}{3}} + \frac{1}{9} =\]

\[= 3^{2} + \frac{1}{9} = 9 + \frac{1}{9} = 9\frac{1}{9}\]

\[4)\ (0,01)^{- 2}\ :100^{- \frac{1}{2}} =\]

\[= \left( \frac{1}{100} \right)^{- 2}\ :\left( (10)^{2} \right)^{- \frac{1}{2}} =\]

\[= 100^{2}\ :10^{- 1} = \left( 10^{2} \right)^{2}\ :\frac{1}{10} =\]

\[= 10^{4} \bullet 10 = 10^{5} = 100\ 000\]

\[5)\ \left( \frac{64}{81} \right)^{- \frac{1}{2}} \bullet \left( \frac{8}{5} \right)^{- 1} = \left( \frac{81}{64} \right)^{\frac{1}{2}} \bullet \frac{5}{8} =\]

\[= \left( \frac{9^{2}}{8^{2}} \right)^{\frac{1}{2}} \bullet \frac{5}{8} = \frac{9}{8} \bullet \frac{5}{8} = \frac{45}{64}\]

\[6)\ \left( 2\frac{10}{27} \right)^{- \frac{2}{3}} \bullet \left( \frac{3}{4} \right)^{2} =\]

\[= \left( \frac{2 \bullet 27 + 10}{27} \right)^{- \frac{2}{3}} \bullet \frac{3^{2}}{4^{2}} =\]

\[= \left( \frac{64}{27} \right)^{- \frac{2}{3}} \bullet \frac{9}{16} = \left( \frac{27}{64} \right)^{\frac{2}{3}} \bullet \frac{9}{16} =\]

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

\[= \frac{9}{16} \bullet \frac{9}{16} = \frac{81}{256}\]