Решебник по алгебре 9 класс Мерзляк Задание 424

\[\boxed{\text{424}\text{\ (424)}\text{.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\left\{ \begin{matrix} x^{2} - x - 6 \leq 0 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} - x - 6 \leq 0\]

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

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

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

\[Ответ:x \in (0;3\rbrack.\]

\[2)\ \left\{ \begin{matrix} 2x^{2} - 11x - 6 \geq 0 \\ x + 4 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ 2x^{2} - 11x - 6 \geq 0\]

\[D = 169\]

\[x_{1,2} = \frac{11 \pm 13}{4}\]

\[x = - 0,5;\ \ \ x = 6\]

\[2)\ x + 4 \geq 0\]

\[x \geq - 4\]

\[\left\{ \begin{matrix} (x + 0,5)(x - 6) \geq 0 \\ x \geq - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x \in \lbrack - 4;\ - 0,5\rbrack \cup \lbrack 6;\ + \infty).\]

\[3)\ \left\{ \begin{matrix} x^{2} - 9x - 10 \leq 0 \\ 6x - x^{2} < 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ x^{2} - 9x - 10 \leq 0\]

\(x_{1} + x_{2} = 9,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 1\)

\[x_{1}x_{2} = - 10,\ \ x_{2} = 10\]

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

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

\[x_{1} = 0,\ \ x_{2} = 6\]

\[\left\{ \begin{matrix} (x + 1)(x - 10) \leq 0 \\ x(6 - x) < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x \in \lbrack - 1;0) \cup (6;10\rbrack.\]

\[4)\ \left\{ \begin{matrix} x^{2} - x - 12 \geq 0 \\ x^{2} + 3x - 10 < 0 \\ \end{matrix} \right.\ \]

\[1)\ x^{2} - x - 12 \geq 0\]

\(x_{1} + x_{2} = 1,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 4\)

\[x_{1}x_{2} = - 12,\ \ x_{2} = - 3\]

\[2)\ x^{2} + 3x - 10 < 0\]

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

\[x_{1}x_{2} = - 10,\ \ x_{2} = 2\]

\[\left\{ \begin{matrix} (x + 3)(x - 4) \geq 0 \\ (x + 5)(x - 2) < 0 \\ \end{matrix} \right.\ \]

\[Ответ:x \in ( - 5; - 3\rbrack.\]