Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 2. П. 5. Решение уравнений Задание 39

\[\boxed{\text{5.\ 39}}\]

\[\textbf{а)}\ x = 6:\]

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

\[3 \cdot 6 = 18\]

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

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

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

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

\[0 \cdot 6 = 0\]

\[2 \cdot 6 = 12\]

\[1 \cdot 6 = 6\]

\[- 6 \cdot 6 = - 36\]

\[\textbf{б)}\ x = \frac{1}{4}:\]

\[\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}\]

\[0 + \frac{1}{4} = \frac{1}{4}\]

\[- 2\frac{1}{4} + \frac{1}{4} = - 2\]

\[0,5 + \frac{1}{4} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}\]

\[\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\]

\[- 1 + \frac{1}{4} = - \frac{3}{4}\]

\[- \frac{1}{2} + \frac{1}{4} = - \frac{2}{4} + \frac{1}{4} = - \frac{1}{4}\]

\[- \frac{3}{4} + \frac{1}{4} = - \frac{2}{4} = - \frac{1}{2}\]

\[\ - \frac{1}{2} + \frac{1}{4} = - \frac{2}{4} + \frac{1}{4} = - \frac{1}{4}\]