Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 2. П. 5. Решение уравнений Задание 122

\[\boxed{\text{5.\ 122}}\]

\[\textbf{а)} - 30 \cdot (x - 21) = - 180\]

\[x - 21 = - 180\ :( - 30)\]

\[x - 21 = 6\]

\[x = 6 + 21\]

\[x = 27.\]

\[\textbf{б)}\ (15 - 9x) \cdot 4 = 204\]

\[15 - 9x = 204\ :4\]

\[15 - 9x = 51\]

\[- 9x = 51 - 15\]

\[- 9x = 36\]

\[x = 36\ :( - 9)\]

\[x = - 4.\]

\[\textbf{в)}\ \frac{9}{14}x - \frac{5}{14} = \frac{1}{7}\ \ \ \ \ | \cdot 14\]

\[9x - 5 = 2\]

\[9x = 2 + 5\]

\[9x = 7\]

\[x = \frac{7}{9}.\]

\[\textbf{г)}\ (3,6 - 0,2x) \cdot 4,9 = 9,8\]

\[3,6 - 0,2x = 9,8\ :4,9\]

\[3,6 - 0,2x = 2\]

\[- 0,2x = 2 - 3,6\]

\[- 0,2x = - 1,6\]

\[x = - 1,6\ :( - 0,2)\]

\[x = 8.\]

\[\textbf{д)}\ (7x - 3,4) \cdot 9 = 13,5\]

\[7x - 3,4 = 13,5\ :9\]

\[7x - 3,4 = 1,5\]

\[7x = 1,5 + 3,4\]

\[7x = 4,9\]

\[x = 4,9\ :7\]

\[x = 0,7.\]

\[\textbf{е)}\ \frac{1}{3}x + \frac{5}{6}x = 3,5\ \ \ \ \ \ | \cdot 6\]

\[2x + 5x = 21\]

\[7x = 21\]

\[x = 21\ :7\]

\[x = 3.\ \]