Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 2. Задачи на повторение 17

\[\boxed{\mathbf{П.\ 17}}\]

\[- \frac{12}{17}\ :\left( - \frac{24}{17} \right) = \frac{12}{17} \cdot \frac{17}{24} = \frac{1}{2} =\]

\[= 0,5\]

\[5,88\ :( - 14,7) = - 58,8\ :147 =\]

\[= - 0,4\]

\[\left( 8 - 5\frac{3}{4} \right) \cdot 2\frac{2}{3} = 2\frac{1}{4} \cdot 2\frac{2}{3} =\]

\[= \frac{9}{4} \cdot \frac{8}{3} = 3 \cdot 2 = 6\]

\[\left( 8 - 6\frac{3}{5} \right)\ :1\frac{3}{4} = 1\frac{2}{5}\ :\frac{7}{4} =\]

\[= \frac{7}{5} \cdot \frac{4}{7} = \frac{4}{5} = 0,8\]

\[\textbf{в)}\ 5,5 - 3\frac{3}{4} \cdot \left( 1\frac{2}{3} + 1\frac{2}{5} \right)\ :2\frac{5}{9} =\]

\[= 5,5 - 4,5 = 1\]

\[1\frac{2}{3} + 1\frac{2}{5} = 1\frac{10}{15} + 1\frac{6}{15} = 2\frac{16}{15} =\]

\[= 3\frac{1}{15}\]

\[3\frac{3}{4} \cdot 3\frac{1}{15}\ :\frac{23}{9} = \frac{15}{4} \cdot \frac{46}{15} \cdot \frac{9}{23} =\]

\[= \frac{2 \cdot 9}{4} = \frac{9}{2} = 4,5\]

\[= 11 - 6,75 = 4,25\]

\[2\frac{4}{5}\ :\frac{7}{5} \cdot 5\frac{1}{2} = \frac{14}{5} \cdot \frac{5}{7} \cdot \frac{11}{2} = 11\]

\[4\frac{2}{7} \cdot \frac{7}{15} \cdot \left( \frac{3}{2} \right)^{3} = \frac{30}{7} \cdot \frac{7}{15} \cdot \frac{27}{8} =\]

\[= \frac{27}{4} = 6\frac{3}{4} = 6,75\]

\[\textbf{д)} - \frac{5}{8} \cdot \frac{4}{15} - \frac{14}{33}\ :\left( - \frac{7}{11} \right) + \frac{1}{12} =\]

\[= - \frac{1}{6} + \frac{1}{12} + \frac{2}{3} =\]

\[= - \frac{2}{12} + \frac{1}{12} + \frac{8}{12} = \frac{7}{12}\]

\[\frac{14}{33} \cdot \left( - \frac{11}{7} \right) = - \frac{2}{3}\]

\[= \frac{2}{7} \cdot \left( \frac{7}{2} \right)^{2} - \frac{5}{13}\ :\frac{40}{13} + \frac{9}{10}\ :\frac{18}{5} =\]

\[= \frac{2}{7} \cdot \frac{49}{4} - \frac{5}{13} \cdot \frac{13}{40} + \frac{9}{10} \cdot \frac{5}{18} =\]

\[= \frac{7}{2} - \frac{1}{8} + \frac{1}{4} = 3\frac{1}{2} - \frac{1}{8} + \frac{2}{8} =\]

\[= 3\frac{4}{8} + \frac{1}{8} = 3\frac{5}{8}\]