Решебник по математике 6 класс Мерзляк ФГОС Задание 382

\[\boxed{\mathbf{382\ (382).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[На\ рисунке\ изображено\ 13\ \]

\[равносторонних\ \]

\[треугольников:9\ маленьких;\]

\[3\ средних;1\ большой.\]

\[\boxed{\mathbf{382}\mathbf{.}}\]

\[1)\ \left( x + \frac{4}{21} \right) - \frac{4}{15} = \frac{16}{35}\]

\[x + \frac{4}{21} = \frac{16^{\backslash 3}}{35} + \frac{4^{\backslash 7}}{15}\]

\[x + \frac{4}{21} = \frac{48 + 28}{105}\]

\[x + \frac{4}{21} = \frac{73}{105}\]

\[x = \frac{73}{105} - \frac{4^{\backslash 5}}{21}\]

\[x = \frac{73}{105} - \frac{20}{105}\]

\[x = \frac{53}{105}.\]

\[2)\ 3\frac{1}{6} - \left( x + 1\frac{1}{12} \right) = \frac{1}{4}\]

\[x + 1\frac{1}{12} = 3\frac{1^{\backslash 2}}{6} - \frac{1^{\backslash 3}}{4}\]

\[x + 1\frac{1}{12} = 3\frac{2}{12} - \frac{3}{12}\]

\[x + 1\frac{1}{12} = 2\frac{14}{12} - \frac{3}{12}\]

\[x + 1\frac{1}{12} = 2\frac{11}{12}\]

\[x = 2\frac{11}{12} - 1\frac{1}{12}\]

\[x = 1\frac{10}{12}\]

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