Решебник по математике 6 класс Ерина Никольский рабочая тетрадь Часть 1, 2 Часть 1 | Страница 77

\[\mathbf{Страница\ 77.}\]

\[\boxed{\mathbf{7.}}\]

\[\textbf{б)}\frac{- 7}{18} = \frac{7}{- 18}\]

\[\textbf{в)} - \frac{3}{8} = \frac{3}{- 8}\]

\[\textbf{г)} - \frac{9}{13} = \frac{9}{- 13}\]

\[\boxed{\mathbf{8.}}\]

\[\textbf{а)}\ да\]

\[\textbf{б)}\ да\]

\[\textbf{в)}\ нет\]

\[\textbf{г)}\ нет\]

\[\textbf{д)}\ да\]

\[\textbf{е)}\ да\]

\[\boxed{\mathbf{9.}}\]

\[\textbf{а)} - \frac{11}{13}\]

\[\textbf{б)} - \frac{3}{11}\]

\[\textbf{в)} - \frac{16}{17}\]

\[\textbf{г)} - \frac{15}{19}\]

\[\textbf{д)} - \frac{5}{9}\]

\[\textbf{е)} - \frac{19}{13}\]

\[\boxed{\mathbf{10.}}\]

\[\textbf{а)}\ 5\frac{14}{23} - 3x = 2\frac{2}{23}\]

\[3x = 5\frac{14}{23} - 2\frac{2}{23}\ \]

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

\[x = \frac{81}{23} \cdot \frac{1}{3}\]

\[x = \frac{27}{23}\]

\[x = 1\frac{4}{23}.\]

\[Ответ:\ \ 1\frac{4}{23}.\]

\[\textbf{б)}\ 5\frac{3}{23} - 2x = 4\frac{12}{23}\]

\[2x = 4\frac{26}{23} - 4\frac{12}{23}\]

\[2x = \frac{14}{23}\]

\[x = \frac{14}{23} \cdot \frac{1}{2}\]

\[x = \frac{7}{23}.\]

\[Ответ:\ \frac{7}{23}.\]

\[\boxed{\mathbf{11.}}\]

\[\textbf{а)}\ \frac{2}{23} \cdot \frac{3}{4} + \frac{4}{5} \cdot \frac{10}{11} = \frac{3}{46} + \frac{8}{11} =\]

\[= \frac{33}{506} + \frac{368}{506} = \frac{401}{506}\]

\[\textbf{б)}\ \frac{10}{11} \cdot \frac{4}{5} + \frac{5}{6} \cdot \frac{12}{11} = \frac{8}{11} + \frac{10}{11} =\]

\[= \frac{18}{11} = 1\frac{7}{11}\]

\[\textbf{в)}\ 2\frac{1}{3} \cdot \left( 14\frac{5}{7} \cdot \frac{3}{7} \right) =\]

\[= \frac{7}{3} \cdot \frac{3}{7} \cdot 14\frac{5}{7} = 14\frac{5}{7}\]

\[\textbf{г)}\ 3\frac{1}{2} \cdot \left( 12\frac{7}{9} \cdot \frac{2}{7} \right) =\]

\[= \frac{7}{2} \cdot \frac{2}{7} \cdot 12\frac{7}{9} = 12\frac{7}{9}\]

\[\textbf{д)}\ \left( \frac{11}{35} \cdot \frac{5}{9} \right) \cdot 1\frac{4}{5} = \frac{11}{35} \cdot \frac{5}{9} \cdot \frac{9}{5} =\]

\[= \frac{11}{35}\ \]

\[\textbf{е)}\ \frac{7}{12} \cdot \frac{6}{11} + \frac{1}{8}\ :\frac{11}{6} =\]

\[= \frac{7}{12} \cdot \frac{6}{11} + \frac{1}{8} \cdot \frac{6}{11} =\]

\[= \frac{6}{11} \cdot \left( \frac{7}{12} + \frac{1}{8} \right) =\]

\[= \frac{6}{11} \cdot \left( \frac{14}{24} + \frac{3}{24} \right) =\]

\[= \frac{6}{11} \cdot \frac{17}{24} = \frac{17}{44}\]

\[\textbf{ж)}\ \left( \frac{13}{78} \cdot \frac{3}{5} \right) \cdot 1\frac{2}{3} = \frac{1}{6} \cdot \frac{3}{5} \cdot \frac{5}{3} = \frac{1}{6}\]

\[\textbf{з)}\ \frac{7}{13} \cdot \frac{5}{14} + \frac{5}{13} \cdot \frac{1}{2} = \frac{5}{26} + \frac{5}{26} =\]

\[= \frac{10}{26} = \frac{5}{13}\]

\[\boxed{\mathbf{12.}}\]

\[1)\ 1\frac{1}{10} \cdot \frac{1}{5} = \frac{11}{10} \cdot \frac{1}{5} = \frac{11}{50}\ (т) -\]

\[меда\ собирают\ пчелы\ с\ \]

\[акации.\]

\[2)\frac{11}{50} \cdot \frac{1}{11} = \frac{1}{50}\ т = \frac{20}{1000}\ т =\]

\[= 20\ (кг) - меда\ собирают\ \]

\[пчелы\ с\ яблони.\]

\[Ответ:20\ кг.\]