Решебник по алгебре 9 класс Мерзляк Задание 405

\[\boxed{\text{405\ (405).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

\[Решение\ квадратных\ \]

\[неравенств.\]

Решение.

\[1)\ x^{2} + 4x + 3 > 0\]

\[x_{1} + x_{2} = - 4,\ \ x_{1} = - 3\]

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

\[Ответ:x \in ( - \infty; - 3) \cup ( - 1;\ + \infty).\]

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

\[x_{1} + x_{2} = 3,\ \ x_{1} = 2\]

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

\[Ответ:x \in \lbrack 1;2\rbrack.\]

\[3) - x^{2} + 12x + 45 < 0\]

\[x_{1} + x_{2} = 12,\ \ x_{1} = 15\]

\[x_{1}x_{2} = - 45,\ \ x_{2} = - 3\]

\[Ответ:x \in ( - \infty; - 3) \cup (15;\ + \infty).\]

\[4) - 3x^{2} - 5x - 2 \geq 0\]

\[D = 25 - 24 = 1\]

\[x = \frac{5 \pm 1}{- 6} = - 1;\ - \frac{2}{3}\]

\[Ответ:x \in \left\lbrack - 1;\ - \frac{2}{3} \right\rbrack.\]

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

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

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

\[Ответ:x \in ( - \infty;0) \cup (5;\ + \infty)\text{.\ }\]

\[6) - 25x^{2} + 16 \leq 0\]

\[x_{1,2} = \pm \frac{4}{5} = \pm 0,8\]

\[Ответ:x \in ( - \infty; - 0,8\rbrack \cup \lbrack 0,8;\ + \infty).\]

\[7)\ 5x^{2} - 3x + 1 \geq 0\]

\[D = 9 - 20 < 0\]

\[Ответ:x \in ( - \infty; + \infty)\text{.\ }\]

\[8) - 3x^{2} + 6x - 4 > 0\]

\[D = 36 - 48 < 0\]

\[Ответ:\ \varnothing.\]

\[9)\ \frac{1}{3}x^{2} - 2x + 3 \leq 0\]

\[D = 4 - 4 = 0\]

\[x = \frac{2}{\frac{2}{3}} = 3\]

\[Ответ:x = 3.\ \]

\[10) - x^{2} + \frac{1}{3}x - \frac{1}{36} > 0\]

\[D = \frac{1}{9} - \frac{1}{9} = 0\]

\[x = \frac{- \frac{1}{3}}{- 2} = \frac{1}{6}\]

\[Ответ:\ \varnothing.\ \]

\[11)\ 2x^{2} - 2x + 0,5 < 0\]

\[D = 4 - 4 = 0\]

\[x = \frac{1}{2}\]

\[Ответ:\ \varnothing.\]