Решебник по алгебре 9 класс Мерзляк Задание 408

\[\boxed{\text{408\ (408).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

\[Решение\ квадратных\ \]

\[неравенств.\]

Решение.

\[1)\ x(x + 5) - 2 < 4x\]

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

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

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

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

\[(x - 1)(x + 2) > 0\]

\[Ответ:x \in ( - 2;1).\]

\[2)\ 11 - (x + 1)^{2} \leq x\]

\[11 - x^{2} - 2x - 1 - x \leq 0\]

\[- x^{2} - 3x + 10 \leq 0\]

\[x^{2} + 3x - 10 \geq 0\]

\[x_{1} + x_{2} = - 3,\ \ x_{1} = - 5\]

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

\[(x + 5)(x - 2) \geq 0\]

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

\[3)\ (2x + 1)^{2} -\]

\[- (x + 1)(x - 7) \leq 5\]

\[4x^{2} + 4x + 1 - x^{2} + 6x + 7 -\]

\[- 5 \leq 0\]

\[3x^{2} + 10x + 3 \leq 0\]

\[D = 100 - 36 = 64\]

\[x_{1,2} = \frac{- 10 \pm 8}{6};\ \ \ \]

\[\ x = - 3;\ \ \ x = - \frac{1}{3}\]

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

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

\[- (2x - 3)(2x + 3) > 30\]

\[5x^{2} + 20x - 4x^{2} + 9 - 30 > 0\]

\[x^{2} + 20x - 21 > 0\]

\[x_{1} + x_{2} = - 20,\ \ x_{1} = - 21\]

\[x_{1}x_{2} = - 21,\ \ x_{2} = 1\]

\[Ответ:x \in ( - \infty;\ - 21) \cup (1;\ + \infty).\]

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

\[- (x - 4)(x + 5) > 30\]

\[3x^{2} - x - 14 - x^{2} - x +\]

\[+ 20 - 30 > 0\]

\[2x^{2} - 2x - 24 > 0\ \ |\ :2\]

\[x^{2} - x - 12 > 0\]

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

\[x_{1}x_{2} = - 12,\ \ x_{2} = - 3\]

\[Ответ:x \in ( - \infty;\ - 3) \cup (4;\ + \infty).\]

\[6)\ \frac{2x^{2} - 1}{4} - \frac{3 - 4x}{6} +\]

\[+ \frac{8x - 5}{8} \leq \frac{19}{24}\ \ \ | \cdot 24\]

\[6 \cdot \left( 2x^{2} - 1 \right) - 4 \cdot (3 - 4x) +\]

\[+ 3 \cdot (8x - 5) - 19 \leq 0\]

\[12x^{2} - 6 - 12 + 16x + 24x -\]

\[- 15 - 19 \leq 0\]

\[12x^{2} + 40x - 52 \leq 0\ \ |\ :4\]

\[3x^{2} + 10x - 13 \leq 0\]

\[D = 100 + 156 = 256\]

\[x_{1,2} = \frac{- 10 \pm 16}{6}\]

\[x = 1;\ \ \ x = - 4\frac{1}{3}\]

\[Ответ:x \in \left\lbrack - 4\frac{1}{3};1 \right\rbrack.\]