Решебник по математике 6 класс Ерина Никольский рабочая тетрадь Часть 1, 2 Часть 1 | Страница 61

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\[\textbf{а)}\ 6\frac{4}{9} \cdot 1\frac{7}{29} - \left( 9\frac{1}{6} - 3\frac{3}{4} \right)\ \ :4\frac{1}{6} =\]

\[= 6,7\]

\[9\frac{1}{6} - 3\frac{3}{4} = 9\frac{2}{12} - 3\frac{9}{12} =\]

\[= 8\frac{14}{12} - 3\frac{9}{12} = = 5\frac{5}{12}\]

\[6\frac{4}{9} \cdot 1\frac{7}{29} = \frac{58}{9} \cdot \frac{36}{29} = 2 \cdot 4 = 8\]

\[5\frac{5}{12}\ :4\frac{1}{6} = \frac{65}{12}\ :\frac{25}{6} = \frac{65}{12} \cdot \frac{6}{25} =\]

\[= \frac{13}{2 \cdot 5} = \frac{13}{10} = 1,3\]

\[8 - 1,3 = 6,7\]

\[\textbf{б)}\ 5\frac{4}{7}\ :2\frac{9}{13} - 1\frac{2}{3}\ :\left( 4\frac{2}{9} - 2\frac{5}{6} \right) =\]

\[= \frac{213}{245}\]

\[4\frac{2}{9} - 2\frac{5}{6} = 4\frac{4}{18} - 2\frac{15}{18} =\]

\[= 3\frac{22}{18} - 2\frac{15}{18} = 1\frac{7}{18}\]

\[5\frac{4}{7}\ :2\frac{9}{13} = \frac{39}{7}\ :\frac{35}{13} =\]

\[= \frac{39}{7} \cdot \frac{13}{35} = \frac{507}{245}\]

\[1\frac{2}{3}\ :1\frac{7}{18} = \frac{5}{3}\ :\frac{25}{18} = \frac{5}{3} \cdot \frac{18}{25} =\]

\[= \frac{90}{75} = \frac{6}{5}\]

\[\frac{507}{245} - \frac{6}{5} = \frac{507}{245} - \frac{294}{245} = \frac{213}{245}\]

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\[\textbf{а)}\ 3\frac{2}{5}\ :\left( 5\frac{3}{14} - x \right) = 2\frac{1}{10}\]

\[5\frac{3}{14} - x = 3\frac{2}{5}\ :2\frac{1}{10}\]

\[5\frac{3}{14} - x = \frac{17}{5} \cdot \frac{10}{21}\]

\[5\frac{3}{14} - x = \frac{34}{21}\]

\[5\frac{3}{14} - x = 1\frac{13}{21}\]

\[x = 5\frac{9}{42} - 1\frac{26}{42}\]

\[x = 4\frac{51}{42} - 1\frac{26}{42}\]

\[x = 3\frac{25}{42}.\]

\[Ответ:3\frac{25}{42}.\]

\[\textbf{б)}\ 8\frac{4}{9} - x = 6\frac{7}{15}\]

\[x = 8\frac{20}{45} - 6\frac{21}{45}\]

\[x = 7\frac{65}{45} - 6\frac{21}{45}\]

\[x = 1\frac{44}{45}.\]

\[Ответ:1\frac{44}{45}\text{.\ }\]