Решебник по алгебре 9 класс Макарычев Задание 330

 

\[\boxed{\text{330\ (330).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 5 \cdot (x - 13)(x + 24) < 0\]

\[5 \cdot (x + 24)(x - 13) < 0\]

\[x \in ( - 24;13).\]

\[\textbf{б)} - \left( x + \frac{1}{7} \right)\left( x + \frac{1}{3} \right) \geq 0\]

\[\left( x + \frac{1}{3} \right)\left( x + \frac{1}{7} \right) \leq 0\]

\[x \in \left\lbrack - \frac{1}{3};\ - \frac{1}{7} \right\rbrack.\]

\[\textbf{в)}\ (x + 12)(3 - x) > 0\]

\[(x + 12)(x - 3) < 0\]

\[x \in ( - 12;3).\]

\[\textbf{г)}\ (6 + x)(3x - 1) \leq 0\]

\[3 \cdot (x + 6)\left( x - \frac{1}{3} \right) \leq 0\]

\[x \in \left\lbrack - 6;\frac{1}{3} \right\rbrack.\]

\[\boxed{\text{330.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{1}{x^{2} - 6x + 8} - \frac{1}{x - 2} + \frac{10}{x^{2} - 4} = 0\]

\[ОДЗ:\ \ \ \ x \neq 2,\ \ x \neq - 2,\]

\[\ \ x \neq 4.\]

\[- x^{2} + 13x - 30 = 0\]

\[x^{2} - 13x + 30 = 0\]

\[По\ теореме\ Виета:\ \ x_{1} = 10,\ \ \]

\[x_{2} = 3.\]

\[\textbf{б)}\ \frac{3}{x^{2} - x - 6} + \frac{3}{x + 2} = \frac{7}{x^{2} - 9}\]

\[ОДЗ:\ \ \ x \neq - 2,\ \ x \neq 3,\]

\[\ \ x \neq - 3.\]

\[3x^{2} + 9 + 3 \cdot \left( x^{2} - 9 \right) - 7x - 14 = 0\]

\[3x^{2} - 4x - 32 = 0\]

\[D = 4 + 3 \cdot 32 = 100\]

\[x_{1,2} = \frac{2 \pm 10}{3} = 4;\ - \frac{8}{3}.\]

\[Ответ:а)\ 3;10;\ \ б) - 2\frac{2}{3};4.\]