Решебник по алгебре 9 класс Мерзляк Задание 434

 

\[\boxed{\text{434\ (434).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ (x - 3)\sqrt{14 + 5x - x^{2}} > 0\]

\[\left\{ \begin{matrix} x - 3 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 14 + 5x - x^{2} > 0 \\ \end{matrix} \right.\ \]

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

\[x > 3\]

\[2)\ 14 + 5x - x^{2} > 0\]

\(x_{1} + x_{2} = 5,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 7\)

\[x_{1}x_{2} = - 14,\ \ x_{2} = - 2\]

\[Ответ:x \in (3;7).\]

\[2)\ (x - 3)\sqrt{14 + 5x - x^{2}} \geq 0\]

\[\left\{ \begin{matrix} x - 3 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 14 + 5x - x^{2} \geq 0 \\ \end{matrix} \right.\ \]

\[1)\ x - 3 \geq 0\]

\[x \geq 3\]

\[2)\ - x^{2} + 5x + 14 \geq 0\]

\[x_{1} = 7,\ \ x_{2} = - 2\]

\[Ответ:x \in \lbrack 3;7\rbrack.\]

\[3)\ (x - 3)\sqrt{14 + 5x - x^{2}} < 0\]

\[\left\{ \begin{matrix} x - 3 < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 14 + 5x - x^{2} > 0 \\ \end{matrix} \right.\ \ \]

\[1)\ x - 3 < 0\]

\[x < 3\]

\[2)\ 14 + 5x - x^{2} > 0\]

\[Ответ:x \in ( - 2;3).\]

\[4)\ (x - 3)\sqrt{14 + 5x - x^{2}} \leq 0\]

\[\left\{ \begin{matrix} x - 3 \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 14 + 5x - x^{2} \geq 0 \\ \end{matrix} \right.\ \]

\[1)\ x - 3 \leq 0\]

\[x \leq 3\]

\[2)\ 14 + 5x - x^{2} \geq 0\]

\[x_{1} = - 2;\ \ \ \ x_{2} = 7\]

\[Ответ:x \in \lbrack - 2;3\rbrack.\]

\[\boxed{\text{434.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x^{2} - 3x - 1 = - \frac{3}{x}\]

\[y = x^{2} - 3x - 1\]

\[x_{0} = \frac{3}{2} = 1,5\]

\[y_{0} = - \frac{13}{4} = - 3\frac{1}{4}\]

\[Oy:\ \ x = 0,\ \ y = - 1\]

\[y = - \frac{3}{x}\]

\[Ответ:\ x = - 1;x = 1;x = 3.\]