Решебник по алгебре 9 класс Мерзляк Задание 425

 

\[\boxed{\text{425\ (425).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\left\{ \begin{matrix} - 6x^{2} + 13x - 5 \leq 0 \\ 6 - 2x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[- 6x^{2} + 13x - 5 \leq 0\]

\[D = 169 - 120 = 49\]

\[x_{1,2} = \frac{- 13 \pm 7}{- 12}\]

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

\[2)\ 6 - 2x > 0\]

\[6 > 2x\]

\[x < 3\]

\[\left\{ \begin{matrix} (x - 0,5)\left( x - 1\frac{2}{3} \right) \leq 0 \\ x < 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x \in ( - \infty;0,5\rbrack \cup \left\lbrack 1\frac{2}{3};3 \right).\]

\[2)\ \left\{ \begin{matrix} x^{2} - 7x - 18 < 0 \\ 5x - x^{2} \leq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ x^{2} - 7x - 18 < 0\]

\(x_{1} + x_{2} = 7,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 9\)

\[x_{1}x_{2} = - 18,\ \ x_{2} = - 2\]

\[2)\ 5x - x^{2} \leq 0\]

\[x(5 - x) \leq 0\]

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

\[\left\{ \begin{matrix} (x - 9)(x + 2) < 0 \\ x(5 - x) \leq 0\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x \in ( - 2;0\rbrack \cup \lbrack 5;9).\]

\[\boxed{\text{425.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[f(x) = x^{2} - 2x - 15\]

\[1)\ x^{2} - 2x - 15 = 0\]

\[x_{1}{+ x}_{2} = 2,\ \ x_{1} = 5\]

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

\[Ответ:\ - 3;5.\]

\[2)\ x^{2} - 2x - 15 = - 7\]

\[x^{2} - 2x - 8 = 0\]

\[x_{1}{+ x}_{2} = 2,\ \ x_{1} = 4\]

\[x_{1}x_{2} = - 8,\ \ x_{2} = - 2\]

\[Ответ:4;\ - 2.\]

\[3)\ x^{2} - 2x - 15 = 33\]

\[x^{2} - 2x - 48 = 0\]

\[x_{1}{+ x}_{2} = 2,\ \ x_{1} = 8\]

\[x_{1}x_{2} = - 48,\ \ x_{2} = - 6\]

\[Ответ:8;\ - 6.\]