Решебник по алгебре 9 класс Макарычев Задание 601

\[\boxed{\text{601\ (601).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ (2x - 1)(x + 8) > 0\]

\[2 \cdot (x - 0,5)(x + 8) > 0\]

\[x \in ( - \infty;\ - 8) \cup (0,5;\ + \infty).\]

\[\textbf{б)}\ (33 - x)(16 + 2x) \leq 0\]

\[2 \cdot (x - 33)(x + 8) \geq 0\]

\[x \in ( - \infty; - 8\rbrack\lbrack 33;\ + \infty).\]

\[\boxed{\text{601.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x_{1} = 2,\ \ x_{4} = x_{1} \cdot q^{3} = \frac{1}{4},\]

\[\ \ 2q^{3} = \frac{1}{4},\ \ q^{3} = \frac{1}{8},\]

\[q = \frac{1}{2},\ \ a = x_{2} = x_{1} \cdot q = 1,\]

\[\ \ b = x_{3} = x_{2} \cdot q = \frac{1}{2}.\]

\[Ответ:a = 1,\ \ b = \frac{1}{2}.\]