Решебник по алгебре 9 класс Мерзляк Задание 130

 

\[\boxed{\text{130\ (130).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[9c - 2 \leq 4c + 4\]

\[9c - 4c \leq 6\]

\[5c \leq 6\]

\[c \leq \frac{6}{5}\]

\[c \leq 1,2\]

\[Ответ:при\ c \in ( - \infty;1,2\rbrack.\]

\[\boxed{\mathbf{130.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1) - 4x - 6 = 7 - 6x\]

\[2x = 13\]

\[x = 6,5.\]

\[2)\ x + 1 = - 4(5 - 4x)\]

\[x + 1 = - 20 + 16x\]

\[- 15x = - 21\]

\[x = \frac{21}{15} = \frac{7}{5} = \frac{14}{10}\]

\[x = 1,4.\]

\[3)\ 3(0,5x - 4) + 8,5x = 18\]

\[1,5x - 12 + 8,5x = 18\]

\[10x = 30\]

\[x = 3.\]

\[4)\ \frac{x + 2}{3} = \frac{3x - 2}{5}\]

\[5(x + 2) = 3(3x - 2)\]

\[5x + 10 = 9x - 6\]

\[- 4x = - 16\]

\[x = 4.\]

\[5)\ \frac{x - 4}{3} + \frac{x}{2} = 5\ \ \ | \cdot 6\]

\[2(x - 4) + 3x = 30\]

\[2x - 8 + 3x = 30\]

\[5x = 38\]

\[x = \frac{38}{5} = 7\frac{3}{5}\]

\[x = 7,6.\]

\[6)\ (0,2x + 8)(9 - 5x) = 0\]

\[0,2 \cdot ( - 5) \cdot (x + 40)\left( x - \frac{9}{5} \right) = 0\]

\[- (x + 40)(x - 1,8) = 0\]

\[x = - 40;\ \ \ x = 1,8.\]