Решебник по алгебре 9 класс Мерзляк Задание 409

\[\boxed{\text{409\ (409).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

\[Решение\ квадратных\ \]

\[неравенств.\]

Решение.

\[1)\ 2 \cdot (x^{2} + 2) \geq x(x + 5)\]

\[2x^{2} + 4 \geq x^{2} + 5x\]

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

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

\[x_{1} + x_{2} = 5,\ \ x_{1} = 4\]

\[x_{1}x_{2} = 4,\ \ x_{2} = 1\]

\[Ответ:x \in ( - \infty;1\rbrack \cup \lbrack 4;\ + \infty).\]

\[2)\ x - (x + 4)(x + 5) > - 5\]

\[x - x^{2} - 9x - 20 + 5 > 0\]

\[- x^{2} - 8x - 15 > 0\]

\[x_{1} + x_{2} = - 8,\ \ x_{1} = - 5\]

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

\[Ответ:x \in ( - 5;\ - 3).\]

\[3)\ (6x - 1)(6x + 1) -\]

\[- (12x - 5)(x + 2) < 7 - 3x\]

\[36x^{2} - 1 - 12x^{2} - 19x +\]

\[+ 10 - 7 + 3x < 0\]

\[24x^{2} - 16x + 2 < 0\ \ |\ :2\]

\[12x^{2} - 8x + 1 < 0\]

\[D = 64 - 48 = 16\]

\[x_{1,2} = \frac{8 \pm 4}{24}\]

\[x = \frac{1}{6};\ \ \ x = 0,5\]

\[Ответ:x \in \left( \frac{1}{6};0,5 \right).\]

\[4)\ \frac{x - 1}{4} - \frac{2x - 3}{2} < \frac{x^{2} + 3x}{8}| \cdot 8\]

\[2 \cdot (x - 1) - 4 \cdot (2x - 3) -\]

\[- \left( x^{2} + 3x \right) < 0\]

\[2x - 2 - 8x + 12 - x^{2} -\]

\[- 3x < 0\]

\[- x^{2} - 9x + 10 < 0\]

\[x_{1} + x_{2} = - 9,\ \ x_{1} = - 10\]

\[x_{1}x_{2} = - 10,\ \ x_{2} = 1\]

\[Ответ:x \in ( - \infty;\ - 10) \cup (1;\ + \infty).\]