Решебник по алгебре 9 класс Мерзляк Задание 139

 

\[\boxed{\text{139\ (139).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\frac{2 - 3x^{\backslash 10}}{4} \geq \frac{1^{\backslash 8}}{5} - \frac{5x + 6^{\backslash 5}}{8}\]

\[\frac{20 - 30x}{40} \geq \frac{8 - 25x - 30}{40}\ | \cdot 40\]

\[20 - 30x \geq - 25x - 22\]

\[- 5x \geq - 42\]

\[x \leq \frac{42}{5}\]

\[x \leq 8\frac{2}{5}\text{\ \ }\]

\[x \in ( - \infty;8,4\rbrack\]

\[x = \left\{ 1;2;3;4;5;6;7;8 \right\}.\]

\[Ответ:8.\]

\[\boxed{\mathbf{139.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{48}{14 - x} - \frac{48}{14 + x} = 1\]

\[\frac{48}{14 - x} - \frac{48}{14 + x} - 1 = 0;\ \ \ \ \ \]

\[x \neq 14;\ \ \ \ x \neq - 14\]

\[x^{2} + 96x - 196 = 0\]

\[x_{1} + x_{2} = - 96,\ \ \]

\[x_{1}x_{2} = - 196,\ \]

\[\ x_{1} = 2,\ \ x_{2} = - 98\]

\[Ответ:\ x = - 98;x = 2.\]

\[2)\ \frac{x - 1}{x + 2} + \frac{x}{x - 2} = \frac{8}{x^{2} - 4}\]

\[(x - 1)(x - 2) + x(x + 2) - 8 =\]

\[= 0\]

\[x^{2} - x - 2x + 2 + x^{2} + 2x - 8 =\]

\[= 0\ \ \]

\[2x^{2} - x - 6 = 0\]

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{1 + 7}{4} = 2\ (не\ подходит);\ \ \ \ \ \ \]

\[x_{2} = \frac{1 - 7}{4} = - \frac{6}{4} = - 1,5\]

\[Ответ:\ x = - 1,5.\]

\[3)\ \frac{4x - 10}{x - 1} + \frac{x + 6}{x + 1} = 4\]

\[\frac{4x - 10}{x - 1} + \frac{x + 6}{x + 1} - 4 = 0;\ \ \ \ \]

\[x \neq 1;\ \ \ \ x \neq - 1\]

\[x^{2} - x - 12 = 0\]

\[x_{1} + x_{2} = 1,\ \ x_{1}x_{2} = - 12,\ \ \]

\[x_{1} = 4,\ \ x_{2} = - 3\]

\[Ответ:\ x = - 3;x = 4.\]

\[4)\ \frac{4}{x^{2} - 10x + 25} - \frac{1}{x + 5} =\]

\[= \frac{10}{x² - 25}\]

\[- x^{2} + 4x + 45 = 0\]

\[x^{2} - 4x - 45 = 0\]

\[x_{1} + x_{2} = 4;\ \ \ \ \ x_{1} \cdot x_{2} = 4 - 5\ \]

\[x_{1} = - 5\ (не\ подходит);\ \ \ \ \]

\[x_{2} = 9\]

\[Ответ:\ x = 9.\]