ГДЗ по алгебре 7 класс Макарычев Задание 6

\[\boxed{\text{6\ (6).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 6\frac{1}{3} - 8 = 6\frac{1}{3} - 7\frac{3}{3} =\]

\[= - \left( 7\frac{3}{3} - 6\frac{1}{3} \right) = - 1\frac{2}{3}\]

\[\textbf{б)} - 2\frac{2}{7} + 4\frac{3}{5} = 4\frac{3^{\backslash 7}}{5} - 2\frac{2^{\backslash 5}}{7} =\]

\[= 4\frac{21}{35} - 2\frac{10}{35} =\]

\[= (4 - 2) + \left( \frac{21 - 10}{35} \right) =\]

\[= 2 + \frac{11}{35} = 2\frac{11}{35}\]

\[\textbf{в)}\ 5\frac{1}{3} - 6\frac{1}{4} = - \left( 6\frac{1^{\backslash 3}}{4} - 5\frac{1^{\backslash 4}}{3} \right) =\]

\[= - \left( 6\frac{3}{12} - 5\frac{4}{12} \right) =\]

\[= - \left( 5\frac{15}{12} - 5\frac{4}{12} \right) = - \frac{11}{12}\]

\[\textbf{г)}\ \frac{3}{8}\ :\left( - \frac{9}{16} \right) = - \left( \frac{3}{8} \cdot \frac{16}{9} \right) =\]

\[= - \frac{3 \cdot 16}{8 \cdot 9} = - \frac{1 \cdot 2}{1 \cdot 3} = - \frac{2}{3}\]

\[\textbf{д)}\ \frac{5}{12} \cdot ( - 6) = - \left( \frac{5 \cdot 6}{12} \right) = - \frac{5}{2} =\]

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

\[\textbf{е)} - 3\frac{2}{9} \cdot 3 = - \left( \frac{29}{9} \cdot 3 \right) =\]

\[= - \frac{29 \cdot 3}{9} = - \frac{29}{3} = - 9\frac{2}{3}\]

\[\textbf{ж)}\ \frac{4}{7} \cdot ( - 49) = - \left( \frac{4 \cdot 49}{7} \right) =\]

\[= - (4 \cdot 7) = - 28\]

\[\textbf{з)} - 16\ :\left( - \frac{4}{9} \right) = 16 \cdot \frac{9}{4} =\]

\[= \frac{16 \cdot 9}{4} = 4 \cdot 9 = 36\]

\[\textbf{и)} - 3\frac{1}{2} \cdot \left( - 1\frac{3}{7} \right) = \frac{7}{2} \cdot \frac{10}{7} =\]

\[= \frac{7 \cdot 10}{2 \cdot 7} = 5\]