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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 404
Задание 404
\[\boxed{\text{404\ (404).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
\[Решение\ квадратных\ \]
\[неравенств.\]
Решение.
\[1)\ x^{2} + 6x - 7 < 0\]
\[x_{1} + x_{2} = - 6,\ \ x_{1} = - 7\]
\[x_{1}x_{2} = - 7,\ \ x_{2} = 1\]
\[Ответ:x \in ( - 7;1).\]
\[2)\ x^{2} - 2x - 48 \geq 0\]
\[x_{1} + x_{2} = 2,\ \ x_{1} = - 6\]
\[x_{1}x_{2} = - 48,\ \ x_{2} = 8\]
\[Ответ:x \in ( - \infty;\ - 6\rbrack \cup \lbrack 8;\ + \infty).\]
\[3) - x^{2} - 6x - 5 > 0\]
\[x_{1} + x_{2} = - 6,\ \ x_{1} = - 5\]
\[x_{1}x_{2} = 5,\ \ x_{2} = - 1\]
\[Ответ:x \in ( - 5;\ - 1).\]
\[4) - x^{2} + 4x - 3 < 0\]
\[x_{1} + x_{2} = 4,\ \ x_{1} = 3\]
\[x_{1}x_{2} = 3,\ \ x_{2} = 1\]
\[Ответ:x \in ( - \infty;1) \cup (3;\ + \infty).\]
\[5)\ 3x^{2} - 7x + 4 \leq 0\]
\[D = 49 - 48 = 0\]
\[x_{1,2} = \frac{7 \pm 1}{6} = 1;1\frac{1}{3}.\]
\[Ответ:x \in \left\lbrack 1;1\frac{1}{3} \right\rbrack\text{.\ }\]
\[6)\ 2x^{2} + 3x + 1 > 0\]
\[D = 9 - 8 = 1\]
\[x_{1,2} = \frac{- 3 \pm 1}{4} = - 1;\ - 0,5.\ \]
\[Ответ:x \in ( - \infty; - 1) \cup ( - 0,5;\ + \infty).\]
\[7)\ 4x^{2} - 12x \leq 0\]
\[4x \cdot (x - 3) \leq 0\]
\[x_{1} = 0,\ \ x_{2} = 3\]
\[Ответ:x \in \lbrack 0;3\rbrack.\]
\[8)\ 4x^{2} - 9 > 0\]
\[x^{2} > \frac{9}{4}\]
\[x_{1,2} = \pm \frac{3}{2}\]
\[Ответ:x \in ( - \infty; - 1,5) \cup (1,5; + \infty).\ \]
\[9)\ x^{2} - 12x + 36 > 0\]
\[(x - 6)^{2} > 0\]
\[x - 6 > 0\]
\[x_{1} = 6\]
\[Ответ:x \in ( - \infty;6) \cup (6;\ + \infty).\]
\[10)\ 4x^{2} - 12x + 9 \geq 0\]
\[(2x - 3)^{2} \geq 0\]
\[2x - 3 \geq 0\]
\[x \geq 1,5\]
\[Ответ:x \in ( - \infty; + \infty).\]
\[11)\ x^{2} + 4x + 4 < 0\]
\[(x + 2)^{2} < 0\]
\[x = - 2.\]
\[Ответ:\ \varnothing.\]
\[12)\ 49x^{2} - 14x + 1 \leq 0\]
\[D = 196 - 196 = 0\]
\[x = \frac{14}{98} = \frac{1}{7}\ \]
\[Ответ:x = \frac{1}{7}.\]
\[13)\ 2x^{2} - x + 3 > 0\]
\[D = 1 - 24 < 0\]
\[Ответ:x \in ( - \infty; + \infty).\]
\[14)\ 3x^{2} - 4x + 5 \leq 0\]
\[D = 16 - 60 < 0\]
\[Ответ:\ \varnothing.\]
\[15) - 4x^{2} + 5x - 7 > 0\]
\[D = 25 - 112 < 0\]
\[Ответ:\ \varnothing.\]
\[16) - 2x^{2} + 3x - 2 \leq 0\]
\[D = 9 - 16 = - 7 < 0\]
\[Ответ:\ x - любое\ число.\]
