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МатематикаРешебник по алгебре 9 класс Макарычев Задание 320
Задание 320
\[\boxed{\text{320\ (320).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} - 2x - 8 < 0 \\ x^{2} - 9 < 0\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x - 4)(x + 2) < 0 \\ (x - 3)(x + 3) < 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 2x - 8 = 0\]
\[D_{1} = 1 + 8 = 9\]
\[x_{1} = 1 + 3 = 4;\]
\[\text{\ \ }x_{2} = 1 - 3 = - 2.\]
\[x \in ( - 2;3).\]
\[\textbf{б)}\ \left\{ \begin{matrix} 2x^{2} - 13x + 6 < 0 \\ x^{2} - 4x > 0\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 2 \cdot (x - 6)(x - 0,5) < 0 \\ x \cdot (x - 4) > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[2x^{2} - 13x + 6 = 0\]
\[D = 169 - 48 = 121\]
\[x_{1} = \frac{13 + 11}{4} = 6;\ \ \]
\[x_{2} = \frac{13 - 11}{4} = 0,5.\]
\[x \in (4;6).\]
\[\textbf{в)}\ \left\{ \begin{matrix} x^{2} - 6x - 16 > 0\ \ \\ x^{2} + 2x - 120 < 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x - 8)(x + 2) > 0\ \ \ \ \\ (x + 12)(x - 10) < 0 \\ \end{matrix} \right.\ \]
\[1)\ x^{2} - 6x - 16 = 0\]
\[D_{1} = 9 + 16 = 25\]
\[x_{1} = 3 + 5 = 8;\ \ x_{2} =\]
\[= 3 - 5 = - 2.\]
\[2)\ x^{2} + 2x - 120 = 0\]
\[D_{1} = 1 + 120 = 121\]
\[x_{1} = - 1 + 11 = 10;\ \ \ \]
\[x_{2} = - 1 - 11 = 12.\]
\[x \in ( - 12;\ - 2) \cup (8;10).\]
\[\textbf{г)}\ \left\{ \begin{matrix} 3x^{2} + x - 2 \leq 0\ \\ x^{2} + 4x - 12 \leq 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 3 \cdot \left( x - \frac{2}{3} \right)(x + 1) \leq 0 \\ (x + 6)(x - 2) \leq 0\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ 3x^{2} + x - 2 = 0\]
\[D = 1 + 24 = 25\]
\[x_{1} = \frac{- 1 + 5}{6} = \frac{4}{6} = \frac{2}{3};\ \]
\[\ x_{2} = \frac{- 1 - 5}{6} = - 1.\]
\[2)\ x^{2} + 4x - 12 = 0\]
\[D_{1} = 4 + 12 = 16\]
\[x_{1} = - 2 + 4 = 2;\]
\[= \ \ x_{2} = - 2 - 4 = - 6.\]
\[x \in \left\lbrack - 1;\ \frac{2}{3} \right\rbrack.\]
\[\textbf{д)}\ \left\{ \begin{matrix} x^{2} + 4x + 15 > 0\ \ (1) \\ x^{2} - 9x + 8 \leq 0\ \ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\ x^{2} + 4x + 15 = 0\ \]
\[D = 16 - 2 \cdot 4 \cdot 15 < 0 \Longrightarrow\]
\[\Longrightarrow x - любое\ число.\]
\[(2)\ x^{2} - 9x + 8 = 0\]
\[x_{1} + x_{2} = 9;\ \ x_{1} \cdot x_{2} = 8\]
\[x_{1} = 1;\ \ x_{2} = 8.\ \]
\[(x - 1)(x - 8) \leq 0\]
\[x \in \lbrack 1;8\rbrack.\]
\[\textbf{е)}\ \left\{ \begin{matrix} 2x^{2} + 5x - 3 < 0\ \ (1) \\ 3x^{2} + x + 11 < 0\ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\ \ 2x^{2} + 5x - 3 = 0\ \]
\[D = 25 + 4 \cdot 2 \cdot 3 = 49\]
\[x_{1} = \frac{- 5 + 7}{4} = 0,5;\ \ \ \]
\[\ x_{2} = \frac{- 5 - 7}{4} = - 3;\]
\[(2)\ 3x^{2} + x + 11 = 0\]
\[D = 1 - 4 \cdot 3 \cdot 11 < 0 \Longrightarrow\]
\[\Longrightarrow решений\ нет.\]
\[Ответ:\ решений\ нет.\]
\[\boxed{\text{320.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[x^{3} + ax + b = 0\]
\[x^{3} = - ax - b\]
\[y = x^{3}\ и\ \ y = - ax - b;\]
\[\textbf{а)}\ a = 0\]
\[Одно\ решение.\]
\[\textbf{б)}\ a > 0\]
\[Одно\ решение.\]
\[\textbf{в)}\ a < 0\]
\[Одно,\ два\ или\ три\ решения.\]
