Решебник по алгебре 9 класс Мерзляк Задание 133

 

\[\boxed{\text{133\ (134).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ \frac{y}{6} - \frac{5y}{4} < 1\ \ \ \ \ \ \ \ \ | \cdot 12\]

\[2y - 15y \leq 12\]

\[- 13y \leq 12\]

\[y \geq - \frac{12}{13}\]

\[Ответ:x \in \left\lbrack - \frac{12}{13};\ + \infty \right).\]

\[2)\ \frac{x}{10} - \frac{x}{5} > - 2\ \ \ \ \ \ \ \ \ \ \ | \cdot 10\]

\[x - 2x > - 20\]

\[- x > - 20\]

\[x < 20\]

\[Ответ:x \in ( - \infty;20)\text{.\ }\]

\[\boxed{\mathbf{133.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{x^{2} - 6x - 7}{x - 1} = 0;\ \ x \neq 1\]

\[x^{2} - 6x - 7 = 0\]

\[D_{1} = 9 + 7 = 16\]

\[x_{1} = 3 + 4 = 7;\]

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

\[Ответ:\ - 1;\ \ 7.\]

\[2)\ \frac{x^{2} + 10x}{x + 4} = \frac{3x - 12}{x + 4};\ \ \ x \neq - 4\]

\[x^{2} + 10x = 3x - 12\]

\[x^{2} + 7x + 12 = 0\]

\[x_{1} + x_{2} = - 7;\ \ x_{1} \cdot x_{2} = 12\]

\[x_{1} = - 3;\ \ \ x_{2} = - 4\ (не\ подходит).\]

\[Ответ:x = - 3.\]

\[3)\ \ \frac{x^{2} - x - 6}{x^{2} - 9} = 0;\ \ \ x \neq \pm 3\]

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

\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = 3\ (не\ подходит);x_{2} = - 2.\]

\[Ответ:x = - 2.\]

\[4)\ \frac{4x + 14}{x - 1} = x^{\backslash x - 1};\ \ \ \ x \neq 1\]

\[4x + 14 = x^{2} - x\]

\[x^{2} - 5x - 14 = 0\]

\[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = - 14\]

\[x_{1} = - 2;\ \ \ x_{2} = 7.\]

\[Ответ:x = - 2;\ \ x = 7.\]