Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 2. П. 5. Решение уравнений Задание 7

\[\boxed{\text{5.\ 7}}\]

\[\textbf{а)}\ \frac{5}{9} + \left( \frac{4}{9} - \frac{7}{11} \right) = \frac{5}{9} + \frac{4}{9} - \frac{7}{11} =\]

\[= \frac{9}{9} - \frac{7}{11} = 1 - \frac{7}{11} = \frac{4}{11}\]

\[\textbf{б)}\ 5\frac{3}{7} + \left( - \frac{3}{7} - \frac{8}{9} \right) =\]

\[= 5\frac{3}{7} - \frac{3}{7} - \frac{8}{9} = 5 - \frac{8}{9} = 4\frac{1}{9}\]

\[\textbf{в)}\ 4,32 + \left( \frac{12}{13} - 3,32 \right) =\]

\[= 4,32 + \frac{12}{13} - 3,32 =\]

\[= 1 + \frac{12}{13} = 1\frac{12}{13}\]

\[\textbf{г)}\ \frac{7}{15} - \left( \frac{4}{15} - \frac{4}{5} \right) =\]

\[= \frac{7}{15} - \frac{4}{15} + \frac{4}{5} = \frac{3}{15} + \frac{4}{5} =\]

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

\[\textbf{д)}\ 6\frac{7}{9} - \left( 3\frac{4}{9} + 2\frac{1}{3} \right) =\]

\[= 6\frac{7}{9} - 3\frac{4}{9} - 2\frac{1}{3} = 3\frac{3}{9} - 2\frac{1}{3} =\]

\[= 3\frac{1}{3} - 2\frac{1}{3} = 1\]

\[\textbf{е)}\ - 9\frac{11}{12} - \left( \frac{1}{4} - \frac{5}{12} \right) =\]

\[= - 9\frac{11}{12} - \frac{1}{4} + \frac{5}{12} =\]

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

\[= - 9\frac{3}{4}\]

\[\textbf{ж)}\ \left( 4\frac{1}{4} - 6\frac{8}{11} \right) + \left( 6,75 - 3\frac{3}{11} \right) =\]

\[= 4,25 - 6\frac{8}{11} + 6,75 - 3\frac{3}{11} =\]

\[= 11 - 9\frac{11}{11} = 11 - 10 = 1\]

\[\textbf{з)}\ \left( 9\frac{7}{18} - 2,7 \right) - \left( 4\frac{1}{18} + 2,3 \right) =\]

\[= 9\frac{7}{18} - 2,7 - 4\frac{1}{18} - 2,3 =\]

\[= 5\frac{6}{18} - 5 = \frac{6}{18} = \frac{1}{3}\ \]