Решебник по алгебре 8 класс Макарычев ФГОС Задание 1251

\[\boxed{\text{1251.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)} - {0,25}^{- 2} \cdot 100 = - \left( \frac{1}{4} \right)^{- 2} \cdot\]

\[\cdot 100 = - 4^{2} \cdot 100 =\]

\[= - 16 \cdot 100 = - 1600\]

\[\textbf{б)}\ 0,01 \cdot ( - 0,5)^{- 3} = 0,01 \cdot\]

\[\cdot \left( - \frac{1}{2} \right)^{- 3} = 0,01 \cdot ( - 2)^{3} =\]

\[= 0,01 \cdot ( - 8) = - 0,08\]

\[\textbf{в)}\ {0,2}^{- 4} \cdot ( - 1,6) = \left( \frac{1}{5} \right)^{- 4} \cdot\]

\[\cdot ( - 1,6) = 5^{4} \cdot ( - 1,6) =\]

\[= 625 \cdot ( - 1,6) = - 1000\]

\[\textbf{г)}\ {0,1}^{- 1} + {1,1}^{0} = \left( \frac{1}{10} \right)^{- 1} +\]

\[+ 1 = 10 + 1 = 11\]

\[\textbf{д)}\ 3\frac{1}{3} \cdot \left( \frac{2}{3} \right)^{- 2} - 0,5 = \frac{10}{3} \cdot\]

\[\cdot \left( \frac{3}{2} \right)^{2} - \frac{5}{10} = \frac{10}{3} \cdot \frac{9}{4} - \frac{5}{10} =\]

\[= \frac{30}{4} - \frac{5}{10} =\]

\[= \frac{150 - 10}{20} = \frac{140}{20} = 7\]

\[\textbf{е)} - 4^{- 1} \cdot 5 + {2,5}^{2} = - \frac{5}{4} +\]

\[+ \left( \frac{5}{2} \right)^{2} = - \frac{5}{4} + \frac{25}{4} = \frac{20}{4} = 5\]

\[\textbf{ж)}\ ( - 0,21)^{3} \cdot ( - 0,1)^{2} = -\]

\[- \left( \frac{21}{100} \right)^{3} \cdot \left( \frac{1}{10} \right)^{2} =\]

\[= - \frac{9261}{1000000} \cdot \frac{1}{100} =\]

\[= - \frac{9261}{100000000} = - 0,00009261\]

\[\textbf{з)} - 6^{- 1} \cdot 36^{2} \cdot \left( \frac{1}{6} \right)^{3} = - \frac{1}{6} \cdot\]

\[\cdot 6^{4} \cdot \frac{1}{6^{3}} = - 1\]

\[\textbf{и)} - ( - 1)^{0} \cdot \left( - \frac{1}{3} \right)^{5} =\]

\[= - 1 \cdot \left( - \frac{1}{3^{5}} \right) = \frac{1}{243}\]