Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 2. П. 5. Решение уравнений Задание 16

\[\boxed{\text{5.\ 16}}\]

\[\textbf{а)}\ x = 6:\]

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

\[\frac{1}{2} \cdot 6 = 3\]

\[- 2 \cdot 6 = - 12\]

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

\[- 3 \cdot 6 = - 18\]

\[- 1\frac{1}{2} \cdot 6 = - \frac{3}{2} \cdot 6 = - 9\]

\[0 \cdot 6 = 0\]

\[- 1 \cdot 6 = - 6\]

\[\frac{2}{3} \cdot 6 = 4\]

\[1 \cdot 6 = 6\]

\[\textbf{б)}\ x = 3:\]

\[6\ :3 = 2\]

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

\[- 1\ :3 = - \frac{1}{3}\]

\[\frac{1}{3}\ :3 = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}\]

\[\frac{1}{6}\ :3 = \frac{1}{6} \cdot \frac{1}{3} = \frac{1}{18}\]

\[- 6\ :3 = - 2\]

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

\[2\ :3 = \frac{2}{3}\]

\[\frac{1}{2}\ :3 = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}\]

\[- 3\ :3 = - 1\]