Решебник по математике 6 класс Ерина Никольский рабочая тетрадь Часть 1, 2 Часть 1 | Страница 68

\[\mathbf{Страница\ 68.}\]

\[\boxed{\mathbf{5.}}\]

\[\textbf{а)}\ |5| + | - 3| = 5 + 3 = 8\ \]

\[\textbf{б)}\ | - 17| + | - 16| = 17 + 16 =\]

\[= 33\]

\[\textbf{в)}\ | - 25| - | - 3| = 25 - 3 = 22\]

\[\textbf{г)}\ | - 27| \cdot \left| \frac{7}{9} \right| = 27 \cdot \frac{7}{9} = 21\]

\[\textbf{д)}\ \left| \frac{4}{5} \right| \cdot \left| \frac{5}{8} \right| = \frac{4}{5} \cdot \frac{5}{8} = \frac{1}{2}\]

\[\textbf{е)}\ | - 34| \cdot \left| \frac{2}{17} \right| = 34 \cdot \frac{2}{17} = 4\]

\[\textbf{ж)}\ | - 125|\ :\left| \frac{5}{13} \right| = 125 \cdot \frac{13}{5} =\]

\[= 25 \cdot 13 = 325\]

\[\textbf{з)}\ \left| \frac{7}{8} \right|\ :\left| 2\frac{1}{4} \right| = \frac{7}{8}\ :\frac{9}{4} = \frac{7}{8} \cdot \frac{4}{9} = \frac{7}{18}\]

\[\boxed{\mathbf{6.}}\]

\[\textbf{а)}\ \frac{3}{5} - \frac{7}{15} = \frac{9}{15} - \frac{7}{15} = \frac{2}{15}\]

\[\frac{2}{15}\ :\frac{2}{3} = \frac{2}{15} \cdot \frac{3}{2} = \frac{1}{5}\]

\[\frac{1}{5} \cdot \frac{7}{8} = \frac{7}{40}\]

\[\frac{7}{40} \cdot 5\frac{5}{7} = \frac{7}{40} \cdot \frac{40}{7} = 1\]

\[\textbf{б)}\ \frac{1}{6} - \frac{1}{4} = \frac{2}{12} - \frac{3}{12} = - \frac{1}{12}\]

\[\frac{3}{8} - \frac{1}{12} = \frac{9}{24} - \frac{2}{24} = \frac{7}{24}\]

\[\frac{7}{24}\ :9\frac{1}{2} = \frac{7}{24}\ :\frac{19}{2} = \frac{7}{24} \cdot \frac{2}{19} =\]

\[= \frac{7}{12 \cdot 19}\]

\[\frac{7}{12 \cdot 19} \cdot 12 = \frac{7}{19}\]

\[\textbf{в)}\ \frac{7}{9} - \frac{1}{3} = \frac{7}{9} - \frac{3}{9} = \frac{4}{9}\]

\[\frac{4}{9} + \frac{1}{6} = \frac{8}{18} + \frac{3}{18} = \frac{11}{18}\]

\[\frac{11}{18}\ :5\frac{1}{2} = \frac{11}{18}\ :\frac{11}{2} = \frac{11}{18} \cdot \frac{2}{11} = \frac{1}{9}\]

\[\frac{1}{9} \cdot 4\frac{1}{2} = \frac{1}{9} \cdot \frac{9}{2} = \frac{1}{2}\]

\[\textbf{г)}\ \frac{1}{5} - \frac{1}{3} = \frac{3}{15} - \frac{5}{15} = - \frac{2}{15}\]

\[- \frac{2}{15} - \frac{3}{10} = - \frac{4}{30} - \frac{9}{30} = - \frac{13}{30}\]

\[- \frac{13}{30}\ :2\frac{1}{3} = - \frac{13}{30}\ :\frac{7}{3} =\]

\[= - \frac{13}{30} \cdot \frac{3}{7} = - \frac{13}{70}\]

\[- \frac{13}{70} \cdot 10 = - \frac{13}{7} = - 1\frac{6}{7}\]

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

\[1)\ 3\frac{1}{8}\ :5 = \frac{25}{8} \cdot \frac{1}{5} = \frac{5}{8}\ (кг) - \ \]

\[металла\ на\ одну\ деталь.\]

\[2)\ 14 \cdot \frac{5}{8} = \frac{35}{4} = 8,75\ (кг) - \ \]

\[металла\ на\ 14\ деталей.\]

\[Ответ:8,75\ кг.\]

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

\[\textbf{а)}\ \frac{36 \cdot 65 \cdot 77}{42 \cdot 117 \cdot 11} = \frac{4 \cdot 65 \cdot 7}{42 \cdot 13} =\]

\[= \frac{4 \cdot 5}{6} = \frac{10}{3} = 3\frac{1}{3}\]

\[\textbf{б)}\ \frac{12 \cdot 35 \cdot 143}{34 \cdot 77 \cdot 39} = \frac{4 \cdot 5 \cdot 143}{34 \cdot 11 \cdot 13} =\]

\[= \frac{2 \cdot 5 \cdot 13}{17 \cdot 13} = \frac{10}{17}\]

\[\textbf{в)}\ \frac{8 \cdot 75 \cdot 77}{63 \cdot 10 \cdot 22} = \frac{8 \cdot 15 \cdot 7}{63 \cdot 2 \cdot 2} = \frac{30}{9} =\]

\[= \frac{10}{3} = 3\frac{1}{3}\]

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

\[27 \cdot 0,06 = 1,62\ (кг) - сахара\ \ \]

\[в\ 27\ кг\ клубники.\]

\[Ответ:1,62\ кг.\]