Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 2. Задачи на повторение 93

\[\boxed{\mathbf{П.\ 93}}\]

\[\textbf{а)}\ 0,8 \cdot (9 + 2x) = 0,5 \cdot (2 - 3x)\]

\[7,2 + 1,6x = 1 - 1,5x\]

\[1,6x + 1,5x = 1 - 7,2\]

\[3,1x = - 7,2\]

\[x = - 7,2\ :3,1\]

\[x = - 2.\]

\[\textbf{б)}\ 0,5 \cdot (x + 3) = 0,8 \cdot (10 - x)\]

\[0,5x + 1,5 = 8 - 0,8x\]

\[0,5x + 0,8x = 8 - 1,5\]

\[1,3x = 6,5\]

\[x = 6,5\ :1,3\]

\[x = 5.\]

\[\textbf{в)}\ 4,2\ :12,6 = z\ :\frac{6}{7}\]

\[12,6z = 4,2 \cdot \frac{6}{7}\]

\[12,6z = 0,6 \cdot 6\]

\[12,6z = 3,6\]

\[z = 3,6\ :12,6 = \frac{36}{126} = \frac{6}{21}\]

\[z = \frac{2}{7}.\]

\[\textbf{г)}\ 10\ :n = 5\frac{5}{7}\ :1\frac{3}{7}\]

\[5\frac{5}{7}n = 10 \cdot 1\frac{3}{7}\]

\[\frac{40}{7}n = 10 \cdot \frac{10}{7}\]

\[n = \frac{100}{7}\ :\frac{40}{7} = \frac{100}{7} \cdot \frac{7}{40} = \frac{10}{4}\]

\[n = 2,5.\]