Решебник по алгебре 9 класс Мерзляк Задание 426

\[\boxed{\text{426\ (426).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\left\{ \begin{matrix} - 2x^{2} - 5x + 18 \geq 0 \\ x^{2} + 4x - 5 \leq 0\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ - 2x^{2} - 5x + 18 \geq 0\]

\[D = 25 + 144 = 169\]

\[x_{1,2} = \frac{5 \pm 13}{- 4}\]

\[x = - 4,5;\ \ \ \ x = 2\]

\[2)\ x^{2} + 4x - 5 \leq 0\]

\(x_{1} + x_{2} = - 4,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 5\)

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

\[\left\{ \begin{matrix} (x + 4,5)(x - 2) \geq 0 \\ (x + 5)(x - 1) \leq 0\ \ \ \\ \end{matrix} \right.\ \]

\[x \in \lbrack - 4,5;1\rbrack\]

\[Ответ:\ x = - 4;\ - 3;\ - 2;\ \]

\[- 1;0;1.\]

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

\[1)\ x^{2} - \left( \sqrt{5} - 3 \right)x - 3\sqrt{5} \leq 0\]

\[D = 5 - 6\sqrt{5} + 9 + 12\sqrt{5} =\]

\[= 6\sqrt{5} + 14\]

\[x_{1,2} = \frac{\sqrt{5} - 3 \pm \sqrt{6\sqrt{5} + 14}}{2}\text{\ \ \ \ \ }\]

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

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

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

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

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

\[x \in \lbrack - 3;\ - 1) \cup (0;2\rbrack\]

\[Ответ:\ x = - 3;\ - 2;\ 1;2.\]