Решебник по математике 6 класс Мерзляк ФГОС Задание 399

 

\(\boxed{\mathbf{399\ (399).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\)

Пояснение.

Решение.

\[18 \cdot \frac{5}{9} = \frac{5 \cdot \mathbf{18}}{\mathbf{9}} = 10\ (кг) - соли\ \]

\[содержится\ в\ рассоле.\]

\[Ответ:10\ кг.\]

\[\boxed{\mathbf{399}\mathbf{.}}\]

\[1)\ 7\frac{3}{10} + \frac{25}{28}x = 8\frac{13}{35}\]

\[\frac{25}{28}x = 8\frac{13^{\backslash 2}}{35} - 7\frac{3^{\backslash 7}}{10}\]

\[\frac{25}{28}x = 8\frac{26}{70} - 7\frac{21}{70}\]

\[\frac{25}{28}x = 1\frac{5}{70}\]

\[\frac{25}{28}x = 1\frac{1}{14}\]

\[x = 1\frac{1}{14}\ :\frac{25}{28}\]

\[x = \frac{15}{14} \cdot \frac{28}{25} = \frac{3 \cdot 2}{1 \cdot 5}\]

\[x = \frac{6}{5} = 1\frac{1}{5}.\]

\[2)\ \frac{3^{\backslash 3}}{8}x + \frac{7^{\backslash 2}}{12}x - \frac{5^{\backslash 4}}{6}x = \frac{9}{32}\]

\[\frac{9 + 14 - 20}{24}x = \frac{9}{32}\]

\[\frac{3}{24}x = \frac{9}{32}\]

\[\frac{1}{8}x = \frac{9}{32}\]

\[x = \frac{9}{32}\ :\frac{1}{8}\]

\[x = \frac{9}{32} \cdot 8\]

\[x = \frac{9}{4} = 2\frac{1}{4}.\]

\[3)\ 2\frac{1}{3}\ :x - 1\frac{1}{6} = 1\frac{5}{9}\]

\[2\frac{1}{3}\ :x = 1\frac{5^{\backslash 2}}{9} + 1\frac{1^{\backslash 3}}{6}\]

\[2\frac{1}{3}\ :x = 1\frac{10}{18} + 1\frac{3}{18}\]

\[2\frac{1}{3}\ :x = 2\frac{13}{18}\]

\[x = 2\frac{1}{3}\ :2\frac{13}{18}\]

\[x = \frac{7}{3}\ :\frac{49}{18}\]

\[x = \frac{7}{3} \cdot \frac{18}{49}\]

\[x = \frac{6}{7}.\]

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

\[x - 1\frac{1}{6} = 2\frac{1}{3}\ :1\frac{5}{9}\]

\[x - 1\frac{1}{6} = \frac{7}{3}\ :\frac{14}{9}\]

\[x - 1\frac{1}{6} = \frac{7}{3} \cdot \frac{9}{14}\]

\[x - 1\frac{1}{6} = \frac{3}{2}\]

\[x - 1\frac{1}{6} = 1\frac{1}{2}\]

\[x = 1\frac{1^{\backslash 3}}{2} + 1\frac{1}{6}\]

\[x = 1\frac{3}{6} + 1\frac{1}{6}\]

\[x = 2\frac{4}{6}\]

\[x = 2\frac{2}{3}.\]