Решебник по алгебре 9 класс Макарычев Задание 389

 

\[\boxed{\text{389\ (389).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left( x^{2} - 16 \right)(x + 17) > 0\]

\[(x + 17)(x + 4)(x - 4) > 0\]

\[x \in ( - 17; - 4) \cup (4; + \infty).\]

\[\textbf{б)}\ \left( x - \frac{2}{3} \right)\left( x^{2} - 121 \right) < 0\]

\[(x - 11)\left( x - \frac{2}{3} \right)(x + 11) < 0\]

\[x \in ( - \infty;\ - 11) \cup \left( \frac{2}{3};11 \right).\]

\[\textbf{в)}\ x^{3} - 25x < 0\]

\[x\left( x^{2} - 25 \right) < 0\]

\[(x + 5)x(x - 5) < 0\]

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

\[\textbf{г)}\ x^{3} - 0,01x > 0\]

\[x\left( x^{2} - 0,01 \right) > 0\]

\[(x + 0,1)x(x - 0,1) > 0\]

\[x \in ( - 0,1;0) \cup (0,1;\ + \infty).\]

\[\textbf{д)}\ \left( x^{2} - 9 \right)\left( x^{2} - 1 \right) > 0\]

\[(x - 3)(x + 3)(x - 1)(x + 1) > 0\]

\[(x + 3)(x + 1)(x - 1)(x - 3) > 0\]

\[x \in ( - \infty; - 3) \cup ( - 1;1) \cup (3; + \infty).\]

\[\textbf{е)}\ \ \left( x^{2} - 15x \right)\left( x^{2} - 36 \right) < 0\]

\[x(x - 15)(x - 6)(x + 6) < 0\]

\[(x + 6)x(x - 6)(x - 15) < 0\]

\[x \in ( - 6;0) \cup (6;15).\]

\[\boxed{\text{389.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[y^{2} + 4y - 4 - 4y = 0\]

\[y^{2} - 4 = 0\]

\[y^{2} = 4\]

\[y = \pm 2.\]

\[\left\{ \begin{matrix} y = 2 \\ x = 2 \\ \end{matrix} \right.\ \ \ \ \ \ или\ \ \ \ \left\{ \begin{matrix} y = - 2 \\ x = - 6 \\ \end{matrix} \right.\ \]

\[Ответ:(2;2);\ \ ( - 6; - 2).\]

\[x^{2} + x - 2x^{2} + 1 - 4x + 4x^{2} - 7 = 0\]

\[3x^{2} - 3x - 6 = 0\ \ \ \ \ |\ :3\]

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

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

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

\[\left\{ \begin{matrix} x = - 1 \\ y = 3\ \ \ \\ \end{matrix} \right.\ \ \ \ или\ \ \ \left\{ \begin{matrix} x = 2\ \ \\ y = - 3 \\ \end{matrix} \right.\ \]

\[Ответ:( - 1;3);\ \ (2; - 3).\]