ГДЗ по алгебре 9 класс Макарычев Задание 567

\[\boxed{\text{567}\text{\ (567)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[a_{n} = n^{2} - n - 20\]

\[n^{2} - n - 20 = 0\]

\[n_{1} + n_{2} = 1;\ \ \ n_{1} \cdot n_{2} = - 20.\]

\[n_{1} = 5,\ \ n_{2} = - 4;\]

\[\Longrightarrow (n - 5)(n + 4) < 0\ \ \ \]

\[при\ \ n \in ( - 4;5).\]

\[a_{1} = 1 - 1 - 20 = - 20\]

\[a_{2} = 4 - 2 - 20 = - 18\]

\[a_{3} = 9 - 3 - 20 = - 14\]

\[a_{4} = 16 - 4 - 20 = - 8.\]

\[\boxed{\text{567.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ (2x - 1)(x + 8) > 0\]

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

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

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

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

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