ГДЗ по алгебре 9 класс Макарычев Задание 387

\[\boxed{\text{387\ (387).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ (3x - 5)(x + 4)(2 - x) = 0\]

\[x_{1} = \frac{5}{3} = 1\frac{2}{3};\ \ x_{2} = - 4;\ \ x_{3} = 2.\ \]

\[\textbf{б)}\ (3x - 5)(x + 4)(2 - x) > 0\]

\[3 \cdot (x + 4)\left( x - \frac{5}{3} \right)(x - 2) < 0\]

\[x \in ( - \infty;\ - 4) \cup \left( 1\frac{2}{3};2 \right).\]

\[\textbf{в)}\ (3x - 5)(x + 4)(2 - x) < 0\]

\[3 \cdot (x + 4)\left( x - \frac{5}{3} \right)(x - 2) > 0\]

\[x \in \left( - 4;1\frac{2}{3} \right) \cup (2; + \infty).\]