Решебник по алгебре 9 класс Макарычев Задание 356

 

\[\boxed{\text{356\ (356).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x^{3} = x\]

\[1)\ y = x\]

\[y = x^{3}\]

\[Исходя\ из\ графика:\]

\[x_{1} = 0,\ \ x_{2} = 0,\ \ \]

\[x_{3} = - 1.\]

\[2)\ Аналитически:\]

\[x^{3} = x\]

\[x^{3} - x = 0\]

\[x\left( x^{2} - 1 \right) = 0\]

\[x(x - 1)(x + 1) = 0\]

\[x_{1} = 0,\ \ x_{2} = 1,\ \ \]

\[x_{3} = - 1.\]

\[Ответ:\ - 1;0;1.\]

\[\boxed{\text{356.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{x - 8}{x + 4} > 0\]

\[(x + 4)(x - 8) > 0;\ \ \ x \neq - 4\]

\[x \in ( - \infty;\ - 4) \cup (8; + \infty).\]

\[\textbf{б)}\ \frac{x + 16}{x - 11} < 0\]

\[(x + 16)(x - 11) < 0;\ \ \ x \neq 11\]

\[x \in ( - 16;11).\]

\[\textbf{в)}\ \frac{x + 1}{3 - x} \geq 0\ \ \]

\[\left\{ \begin{matrix} (x + 1)(x - 3) \leq 0 \\ x \neq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x \in \lbrack - 1;3).\]

\[\textbf{г)}\ \frac{6 - x}{x - 4} \leq 0\]

\[\left\{ \begin{matrix} (x - 4)(x - 6) \geq 0 \\ x \neq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x \in ( - \infty;4) \cup \lbrack 6;\ + \infty).\]

\[\textbf{д)}\ \frac{2x - 4}{3x + 3} \leq 0\ \ \]

\[\left\{ \begin{matrix} (2x - 4)(3x + 3) \leq 0 \\ 3x \neq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2 \cdot (x + 1)(x - 2) \leq 0 \\ x \neq - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x \in ( - 1;2\rbrack.\]

\[\textbf{е)}\ \frac{5x - 1}{2x + 3} \geq 0\ \ \]

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

\[\left\{ \begin{matrix} 10 \cdot (x + 1,5)(x - 0,2) \geq 0\ \\ x \neq - 1,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x \in ( - \infty; - 1,5)\lbrack 0,2;\ + \infty).\]