ГДЗ по алгебре 9 класс Макарычев Задание 573

\[\boxed{\text{573}\text{\ (573)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x² + x - 42 \leq 0\]

\[x^{2} + x - 42 = 0\]

\[x_{1} + x_{2} = - 1;\ \ \ \ \ x_{1} \cdot x_{2} = - 42\]

\[x_{1} = - 7;\ \ x_{2} = 6.\]

\[(x + 7)(x - 6) \leq 0\]

\[- 7 \leq x \leq 6.\]

\[\textbf{б)}\ (x + 11)(x + 4)(x - 1) > 0\]

\[x = - 11;\ \ x = - 4;\ \ x = 1\]

\[x \in ( - 11;\ - 4) \cup (1;\ + \infty).\]

\[\boxed{\text{573.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[S_{n} = \frac{\left( a_{1} + a_{n} \right)}{2} \cdot n,\]

\[a_{n} = 3n + 2;\ \ a_{1} = 3 + 2 = 5:\ \]

\[\textbf{а)}\ a_{20} = 60 + 2 = 62\ \ \]

\[S_{20} = \frac{5 + 62}{2} \cdot 20 = 67 \cdot 10 =\]

\[= 670.\]

\[\textbf{б)}\ a_{15} = 45 + 2 = 47\]

\[S_{15} = \frac{5 + 47}{2} \cdot 15 = \frac{52}{2} \cdot 15 =\]

\[= 26 \cdot 15 = 390.\]