Решебник по алгебре 9 класс Макарычев Задание 357

 

\[\boxed{\text{357\ (357).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x^{3} + ax + b = 0\]

\[x^{3} = - ax - b\]

\[y = x^{3}\ и\ \ y = - ax - b;\]

\[\textbf{а)}\ a = 0\]

\[Одно\ решение.\]

\[\textbf{б)}\ a > 0\]

\[Одно\ решение.\]

\[\textbf{в)}\ a < 0\]

\[Одно,\ два\ или\ три\ решения.\]

\[\boxed{\text{357.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{6x + 2}{x + 4} < 5\]

\[\frac{6x + 2 - 5 \cdot (x + 4)}{x + 4} < 0\]

\[\frac{6x + 2 - 5x - 20}{x + 4} < 0\]

\[\frac{x - 18}{x + 4} < 0 \Longrightarrow\]

\[\Longrightarrow (x - 18)(x + 4) < 0\]

\[x \in ( - 4;\ 18).\]

\[\textbf{б)}\ \frac{5x + 8}{x} > 1\]

\[\frac{5x + 8 - x}{x} > 0\]

\[\frac{4x + 8}{x} > 0\]

\[x(4x + 8) > 0\]

\[x(x + 2) > 0\]

\[x \in ( - \infty; - 2) \cup (0; + \infty).\]

\[\textbf{в)}\ \frac{3 - 2x}{3x + 2} \leq 1\]

\[\frac{3 - 2x - 3x - 2}{3x + 2} \leq 0\]

\[\frac{1 - 5x}{3x + 2} \leq 0\]

\[\frac{5x - 1}{3x + 2} \geq 0 \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (5x - 1)(3x + 2) \geq 0 \\ x \neq - \frac{2}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[x \in \left( - \infty;\ - \frac{2}{3} \right) \cup \lbrack 0,2; + \infty).\]

\[\textbf{г)}\ \frac{5x - 4}{x + 8} \geq 15\]

\[\frac{5x - 4 - 15 \cdot (x + 8)}{x + 8} \geq 0\]

\[\frac{- 10x - 124}{x + 8} \geq 0\]

\[\frac{x + 12,4}{x + 8} \geq 0 \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x + 12,4)(x + 8) \geq 0 \\ x \neq - 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x \in \lbrack - 12,4;\ - 8).\]