ГДЗ по алгебре 9 класс Макарычев Задание 389

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

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

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

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

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

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

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

\[\textbf{в)}\ x^{3} - 25x < 0\]

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

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

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

\[\textbf{г)}\ x^{3} - 0,01x > 0\]

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

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

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

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

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

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

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

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

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

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

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