Решебник по математике 6 класс Мерзляк ФГОС Задание 1352

 

\[\boxed{\mathbf{1352}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[a = \left( - 3,8 - ( - 4,3) \right)\ :\left( - 1\frac{1}{3} \right) =\]

\[= ( - 3,8 + 4,3)\ :\left( - \frac{4}{3} \right) =\]

\[= 0,5 \cdot \left( - \frac{3}{4} \right) =\]

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

\[= - 0,375 \Longrightarrow |a| = 0,375.\]

\[b = \left( 5\frac{7}{8} - 6\frac{1}{12} \right)\ :1\frac{7}{18} =\]

\[= \left( 5\frac{21}{24} - 6\frac{2}{24} \right)\ :\frac{25}{18} =\]

\[= \left( 5\frac{21}{24} - 5\frac{26}{24} \right) \cdot \frac{18}{25} =\]

\[= - \frac{5}{24} \cdot \frac{18}{25} = - \frac{1 \cdot 3}{4 \cdot 5} = - \frac{3}{20} =\]

\[= - 0,15 \Longrightarrow |b| = 0,15\]

\[c = \left( - 1\frac{5}{8} - ( - 2,15) \right)\ :\]

\[:\left( - 2\frac{4}{5} \right) =\]

\[= \left( - 1\frac{5}{8} + 2\frac{15}{100} \right)\ :\left( - \frac{14}{5} \right) =\]

\[= \left( - 1\frac{5}{8} + 2\frac{3}{20} \right) \cdot \left( - \frac{5}{14} \right) =\]

\[= \left( - 1\frac{25}{40} + 2\frac{6}{40} \right) \cdot \left( - \frac{5}{14} \right) =\]

\[= \left( - 1\frac{25}{40} + 1\frac{46}{40} \right) \cdot \left( - \frac{5}{14} \right) =\]

\[= \frac{21}{40} \cdot \left( - \frac{5}{14} \right) = - \frac{21 \cdot 5}{40 \cdot 14} =\]

\[= - \frac{3}{8 \cdot 2} = - \frac{3}{16} =\]

\[= - 0,1875 \Longrightarrow |c| = 0,1875\]

\[d = \left( - 1\frac{5}{12} - 1\frac{2}{15} \right) \cdot \left( - \frac{5}{17} \right) =\]

\[= \left( - 1\frac{25}{60} - 1\frac{8}{60} \right) \cdot \left( - \frac{5}{17} \right) =\]

\[= - 2\frac{33}{60} \cdot \left( - \frac{5}{17} \right) =\]

\[= 2\frac{11}{20} \cdot \frac{5}{17} = \frac{51 \cdot 5}{20 \cdot 17} = \frac{3}{4} =\]

\[= 0,75 \Longrightarrow |d| = 0,75\]

\[Наибольший\ модуль\ у\ числа\ a:\]

\[\text{\ \ }|d| = 0,75.\ \]

\[Ответ:d = 0,75.\]