Решебник по алгебре 9 класс Мерзляк Задание 433

 

\[\boxed{\text{433\ (433).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)(x + 4)\sqrt{x^{2} - 2x - 15} > 0\]

\[\left\{ \begin{matrix} x + 4 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 2x - 15 > 0 \\ \end{matrix} \right.\ \]

\[1)\ x + 4 > 0\]

\[x > - 4\]

\[2)\ x^{2} - 2x - 15 > 0\]

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

\[x_{1}x_{2} = - 15,\ \ x_{2} = - 3\]

\[Ответ:x \in ( - 4;\ - 3) \cup (5;\ + \infty).\]

\[2)\ (x + 4)\sqrt{x^{2} - 2x - 15} \geq 0\]

\[\left\{ \begin{matrix} x + 4 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 2x - 15 \geq 0 \\ \end{matrix} \right.\ \]

\[1)\ x + 4 \geq 0\]

\[x \geq - 4\]

\[2)\ x^{2} - 2x - 15 \geq 0\]

\[x_{1} = 5,\ \ x_{2} = - 3\]

\[Ответ:x \in \lbrack - 4;\ - 3\rbrack \cup \lbrack 5;\ + \infty).\]

\[3)\ (x + 4)\sqrt{x^{2} - 2x - 15} < 0\]

\[\left\{ \begin{matrix} x^{2} - 2x - 15 > 0 \\ x + 4 < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ x + 4 < 0\]

\[x < - 4\]

\[2)\ x^{2} - 2x - 15 > 0\]

\[x_{1} = 5,\ \ x_{2} = - 3\]

\[Ответ:x \in ( - \infty;\ - 4).\]

\[4)\ (x + 4)\sqrt{x^{2} - 2x - 15} \leq 0\]

\[\left\{ \begin{matrix} x + 4 \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 2x - 15 \geq 0 \\ \end{matrix} \right.\ \]

\[1)\ x + 4 \leq 0\]

\[x \leq - 4\]

\[2)\ x^{2} - 2x - 15 \geq 0\]

\[x_{1} = 5,\ \ x_{2} = - 3\]

\[Ответ:x \in ( - \infty;\ - 4\rbrack.\]

\[\boxed{\text{433.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[f(x) = 3x^{2} - 6x\]

\[x_{0} = \frac{6}{6} = 1\]

\[y_{0} = 3 - 6 = - 3\]

\[Ox:\ \ \ 3x(x - 2) = 0\]

\[x = 0,\ \ (0;0)\]

\[x = 2,\ \ (2;0)\]

\[\text{Oy}:\ \ x = 0,\ \ y = 0\]

\[f( - 1) = 3 + 6 = 9\]

\[1)\ E(y) = \lbrack - 3; + \infty)\]

\[2)\ ( - \infty;1\rbrack\]

\[3)\ x \in ( - \infty;0\rbrack \cup \lbrack 2;\ + \infty)\]