ГДЗ по алгебре 9 класс Мерзляк Задание 432

\[\boxed{\text{432\ (432).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ |x| \cdot \left( x^{2} - 5x + 6 \right) > 0\]

\[\left\{ \begin{matrix} |x| > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5x + 6 > 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 5x + 6 > 0\]

\(x_{1} + x_{2} = 5,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 3\)

\[x_{1}x_{2} = 6,\ \ x_{2} = 2\]

\[Ответ:x \in ( - \infty;0) \cup (0;2) \cup\]

\[\cup (3; + \infty)\ .\]

\[2)\ \sqrt{x} \cdot \left( x^{2} + 6x - 40 \right) > 0\]

\[\left\{ \begin{matrix} x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 6x - 40 > 0 \\ \end{matrix} \right.\ \]

\[x^{2} + 6x - 40 > 0\ \]

\(x_{1} + x_{2} = - 6,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 10\)

\[x_{1}x_{2} = - 40,\ \ x_{2} = 4\]

\[Ответ:x \in (4;\ + \infty).\]

\[3)\ (x + 3)^{2}\left( x^{2} - x - 6 \right) > 0\]

\[\left\{ \begin{matrix} (x + 3)^{2} > 0\ \ \ \ \ \\ x^{2} - x - 6 > 0 \\ \end{matrix} \right.\ \]

\[1)\ (x + 3)^{2} > 0\]

\[x = - 3\]

\[2)\ x^{2} - x - 6 > 0\]

\(x_{1} + x_{2} = 1,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 3\)

\[x_{1}x_{2} = - 6,\ \ x_{2} = - 2\]

\[Ответ:x \in ( - \infty; - 3) \cup\]

\[\cup ( - 3;\ - 2) \cup (3;\ + \infty).\]

\[4)\ \frac{3x^{2} - 8x - 3}{(x - 1)^{2}} \leq 0\]

\[\left\{ \begin{matrix} 3x^{2} - 8x - 3 \leq 0 \\ (x - 1)^{2} > 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ 3x^{2} - 8x - 3 \leq 0\]

\[D = 64 + 36 = 100\]

\[x_{1,2} = \frac{8 \pm 10}{6}\]

\[x = - \frac{1}{3};\ \ \ \ x = 3\]

\[2)\ (x - 1)^{2} > 0\]

\[x = 1\]

\[Ответ:x \in \left\lbrack - \frac{1}{3};1 \right) \cup (1;3\rbrack.\]