ГДЗ по алгебре 9 класс Макарычев Задание 391

\[\boxed{\text{391\ (391).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y = \frac{4}{\sqrt{(3x - 1)(6x + 1)}}\]

\[(3x - 1)(6x + 1) > 0\]

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

\[x \in \left( - \infty; - \frac{1}{6} \right) \cup \left( \frac{1}{3}; + \infty \right).\]

\[\textbf{б)}\ y = \frac{7}{\sqrt{(11x + 2)(x - 4)}}\]

\[(11x + 2)(x - 4) > 0\]

\[11 \cdot \left( x + \frac{2}{11} \right)(x - 4) > 0\]

\[x \in \left( - \infty; - \frac{2}{11} \right) \cup (4; + \infty)\text{.\ }\]