Решебник по математике 6 класс Ерина Виленкин рабочая тетрадь Часть 1, 2 Часть 1 | Страница 96

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\[\textbf{а)}\ \left( \frac{55}{84}\ :x + 1\frac{1}{2} \right) \cdot \frac{5}{33} = 2\frac{1}{2}\]

\[\frac{55}{84}\ :x = 2\frac{1}{2}\ :\frac{5}{33} - 1\frac{1}{2}\]

\[\frac{55}{84}\ :x = \frac{5}{2} \cdot \frac{33}{5} - \frac{3}{2}\]

\[\frac{55}{84}\ :x = \frac{33}{2} - \frac{3}{2}\]

\[\frac{55}{84}\ :x = \frac{30}{2} = 15\]

\[x = \frac{55}{84}\ :15\]

\[x = \frac{11}{252}\]

\[Ответ:\frac{11}{252}.\]

\[\textbf{б)}\ 1\frac{5}{28}\ :\left( x\ :3\frac{3}{5} - \frac{1}{7} \right) = 2\frac{5}{14}\]

\[x\ :3\frac{3}{5} = 2\frac{5}{14}\ :1\frac{5}{28} + \frac{1}{7}\]

\[x\ :3\frac{3}{5} = \frac{33}{14} \cdot \frac{28}{33} + \frac{1}{7}\]

\[x\ :3\frac{3}{5} = 2 + \frac{1}{7}\]

\[x\ :3\frac{3}{5} = 2\frac{1}{7}\]

\[x = 2\frac{1}{7} \cdot 3\frac{3}{5}\]

\[x = \frac{15}{7} \cdot \frac{18}{5}\]

\[x = \frac{54}{7} = 7\frac{5}{7}\]

\[Ответ:7\frac{5}{7}.\]

\[\textbf{в)}\ \left( 1\frac{7}{10}\ :(23 - x) \right) \cdot 21\frac{1}{4} = 2\frac{5}{6}\ \]

\[1\frac{7}{10}\ :(23 - x) = 2\frac{5}{6}\ :21\frac{1}{4}\]

\[1\frac{7}{10}\ :(23 - x) = \frac{17}{6} \cdot \frac{4}{85}\]

\[1\frac{7}{10}\ :(23 - x) = \frac{2}{15}\]

\[23 - x = 1\frac{7}{10} \cdot \frac{2}{15}\]

\[23 - x = \frac{17}{10} \cdot \frac{15}{2}\]

\[23 - x = \frac{51}{4}\]

\[23 - x = 12\frac{3}{4}\]

\[x = 23 - 12\frac{3}{4} = 22\frac{4}{4} - 12\frac{3}{4}\]

\[x = 10\frac{1}{4}\]

\[Ответ:10\frac{1}{4}.\]