Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 469

\[\boxed{\mathbf{2.469}}\]

\[\textbf{а)}\ 1\frac{5}{7}\ :x = \frac{6}{7}\ :2\]

\[\frac{6}{7}x = \frac{12}{7} \cdot 2\]

\[\frac{6}{7}x = \frac{24}{7}\]

\[x = \frac{24}{7}\ :\frac{6}{7} = \frac{24}{7} \cdot \frac{7}{6}\]

\[x = 4\]

\[Ответ:x = 4.\]

\[\textbf{б)}\ a\ :1\frac{3}{4} = 1\frac{3}{4} \cdot \frac{1}{4}\]

\[a\ :\frac{7}{4} = \frac{7}{4} \cdot \frac{1}{4}\]

\[a = \frac{7}{16} \cdot \frac{7}{4}\]

\[a = \frac{49}{64}\]

\[Ответ:a = \frac{49}{64}.\]

\[\textbf{в)}\ 1\frac{2}{3} \cdot \left( \frac{1}{3}n + \frac{3}{7} \right) = 2\frac{1}{4}\]

\[\frac{1}{3}n + \frac{3}{7} = \frac{9}{4}\ :\frac{5}{3}\]

\[\frac{1}{3}n + \frac{3}{7} = \frac{9}{4} \cdot \frac{3}{5}\]

\[\frac{1}{3}n = \frac{27}{20} - \frac{3}{7}\]

\[\frac{1}{3}n = \frac{189}{140} - \frac{60}{140} = \frac{129}{140}\]

\[n = \frac{129}{140}\ :\frac{1}{3} = \frac{129}{140} \cdot 3\]

\[n = \frac{387}{140} = 2\frac{107}{140}\]

\[Ответ:n = 2\frac{107}{140}.\]

\[\textbf{г)}\ \left( \frac{5}{4}z - \frac{3}{5} \right) \cdot \frac{7}{8} = \frac{7}{8}\]

\[\frac{5}{4}z - \frac{3}{5} = 1\]

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

\[\frac{5}{4}z = 1\frac{3}{5}\]

\[z = \frac{8}{5}\ :\frac{5}{4} = \frac{8}{5} \cdot \frac{4}{5}\]

\[z = \frac{32}{25} = 1\frac{7}{25}\]

\[Ответ:z = 1\frac{7}{25}\text{.\ }\]