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МатематикаРешебник по алгебре 9 класс Макарычев Задание 809
Задание 809
\[\boxed{\text{809\ (809).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[0 < P(A) < 1;\]
\[0 < P(B) < 1;\]
\[P(C) = 1;\]
\[P(D) = 0.\]
\(\boxed{\text{809.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)
\[\textbf{а)}\ x² + 2x - 15 < 0\]
\[y = x^{2} + 2x - 15 \Longrightarrow\]
\[\Longrightarrow \mathbf{ветви\ вверх.}\]
\[D = 4 + 60 = 64\]
\[x_{1} = \frac{- 2 + 8}{2} = 3,\ \ \]
\[x_{2} = \frac{- 2 - 8}{2} = - 5,\]
\[\Longrightarrow x \in ( - 5;3).\]
\[\textbf{б)}\ 5x² - 11x + 2 \geq 0\]
\[y = 5x^{2} - 11x + 2 \Longrightarrow\]
\[\Longrightarrow ветви\ вверх.\]
\[D = 121 - 40 = 81,\]
\[x_{1} = \frac{11 + 9}{10} = 2,\ \ \]
\[x_{2} = \frac{11 - 9}{10} = 0,2.\]
\[x \in ( - \infty;0,2\rbrack \cup \lbrack 2; + \infty).\]
\[\textbf{в)}\ 10 - 3x^{2} \leq 5x - 2\]
\[3x² + 5x - 12 \geq 0\]
\[y = 3x^{2} + 5x - 12 \Longrightarrow\]
\[\Longrightarrow ветви\ вверх.\]
\[D = 25 + 144 = 169\]
\[x_{1} = \frac{- 5 + 13}{6} = \frac{8}{6} = 1\frac{1}{3},\ \ \]
\[x_{2} = \frac{- 5 - 13}{6} = - 3,\]
\[x \in ( - \infty; - 3\rbrack \cup \left\lbrack 1\frac{1}{3};\ + \infty \right).\]
\[\textbf{г)}\ (2x + 3)(2 - x) > 3\]
\[4x - 2x^{2} + 6 - 3x - 3 > 0\]
\[- 2x^{2} + x - 3 > 0\]
\[2x^{2} - x - 3 < 0\]
\[y = 2x^{2} - x - 3 \Longrightarrow\]
\[\Longrightarrow ветви\ вверх.\]
\[D = 1 + 24 = 25,\]
\[x_{1} = \frac{1 + 5}{4} = 1,5,\ \ \]
\[x_{2} = \frac{1 - 5}{4} = - 1,\]
\[x \in ( - 1;1,5).\]
\[\textbf{д)}\ 2x² - 0,5 \leq 0\]
\[2x^{2} \leq \frac{1}{2}\]
\[- \frac{1}{2} \leq x \leq \frac{1}{2}.\]
\[\textbf{е)}\ 3x² + 3,6x > 0\]
\[y = 3x^{2} + 3,6x \Longrightarrow ветви\ вверх.\]
\[3x^{2} + 3,6x = 0\]
\[3x(x + 1,2) = 0\]
\[x_{1} = 0,\ \ x_{2} = - 1,2.\]
\[x \in ( - \infty; - 1,2) \cup (0; + \infty).\]
\[\textbf{ж)}\ (0,2 - x)(0,2 + x) < 0\]
\[(x - 0,2)(x + 0,2) > 0\]
\[x^{2} - 0,04 > 0\]
\[x^{2} > 0,04\]
\[x \in ( - \infty; - 0,2) \cup (0,2; + \infty).\]
\[\textbf{з)}\ x(3x - 2,4) > 0\]
\[x(x - 0,8) > 0\]
\[x^{2} - 0,08x > 0\]
\[y = x^{2} - 0,8x \Longrightarrow ветви\ вверх.\]
\[x^{2} - 0,8x = 0\]
\[x_{1} = 0,\ \ x_{2} = 0,8.\]
\[x \in ( - \infty;0) \cup (0,8;\ + \infty).\]
\[\textbf{и)}\ x^{2} - 0,5x - 5 < 0\ \ \ \ \ | \cdot 2\]
\[2x^{2} - x - 10 < 0\]
\[D = 1 + 80 = 81\]
\[x_{1} = \frac{1 + 9}{4} = 2,5;\ \]
\[\ x_{2} = \frac{1 - 9}{4} = - 2.\]
\[(x + 2)(x - 2,5) < 0\]
\[x \in ( - 2;2,5).\]
\[к)\ x^{2} - 2x + 12,5 > 0\]
\[D_{1} = 1 - 12,5 < 0\]
\[нет\ корней.\]
\[x \in ( - \infty; + \infty).\]
