Решебник по алгебре 9 класс Макарычев Задание 1011

 

\[\boxed{\text{1011\ (1011).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x² + 2x - 15 < 0\]

\[y = x^{2} + 2x - 15 \Longrightarrow \mathbf{ветви\ }\]

\[\mathbf{вверх.}\]

\[D = 4 + 60 = 64\]

\[x_{1} = \frac{- 2 + 8}{2} = 3,\ \ \]

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

\[\Longrightarrow x \in ( - 5;3).\]

\[\textbf{б)}\ 5x² - 11x + 2 \geq 0\]

\[y = 5x^{2} - 11x + 2 \Longrightarrow ветви\ \]

\[вверх.\]

\[D = 121 - 40 = 81,\]

\[x_{1} = \frac{11 + 9}{10} = 2,\ \ \]

\[x_{2} = \frac{11 - 9}{10} = 0,2.\]

\[x \in ( - \infty;0,2\rbrack \cup \lbrack 2; + \infty).\]

\[\textbf{в)}\ 10 - 3x^{2} \leq 5x - 2\]

\[3x² + 5x - 12 \geq 0\]

\[y = 3x^{2} + 5x - 12 \Longrightarrow ветви\ \]

\[вверх.\]

\[D = 25 + 144 = 169\]

\[x_{1} = \frac{- 5 + 13}{6} = \frac{8}{6} = 1\frac{1}{3},\ \ \]

\[x_{2} = \frac{- 5 - 13}{6} = - 3,\]

\[x \in ( - \infty; - 3\rbrack \cup \left\lbrack 1\frac{1}{3};\ + \infty \right).\]

\[\textbf{г)}\ (2x + 3)(2 - x) > 3\]

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

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

\[2x^{2} - x - 3 < 0\]

\[y = 2x^{2} - x - 3 \Longrightarrow ветви\ вверх.\]

\[D = 1 + 24 = 25,\]

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

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

\[x \in ( - 1;1,5).\]

\[\textbf{д)}\ 2x² - 0,5 \leq 0\]

\[2x^{2} \leq \frac{1}{2}\]

\[- \frac{1}{2} \leq x \leq \frac{1}{2}.\]

\[\textbf{е)}\ 3x² + 3,6x > 0\]

\[y = 3x^{2} + 3,6x \Longrightarrow ветви\ вверх.\]

\[3x^{2} + 3,6x = 0\]

\[3x(x + 1,2) = 0\]

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

\[x \in ( - \infty; - 1,2) \cup (0; + \infty).\]

\[\textbf{ж)}\ (0,2 - x)(0,2 + x) < 0\]

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

\[x^{2} - 0,04 > 0\]

\[x^{2} > 0,04\]

\[x \in ( - \infty; - 0,2) \cup (0,2; + \infty).\]

\[\textbf{з)}\ x(3x - 2,4) > 0\]

\[x(x - 0,8) > 0\]

\[x^{2} - 0,08x > 0\]

\[y = x^{2} - 0,8x \Longrightarrow ветви\ вверх.\]

\[x^{2} - 0,8x = 0\]

\[x_{1} = 0,\ \ x_{2} = 0,8.\]

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