ГДЗ по алгебре 9 класс Макарычев Задание 320

\[\boxed{\text{320\ (320).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} - 2x - 8 < 0 \\ x^{2} - 9 < 0\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x - 4)(x + 2) < 0 \\ (x - 3)(x + 3) < 0 \\ \end{matrix} \right.\ \]

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

\[D_{1} = 1 + 8 = 9\]

\[x_{1} = 1 + 3 = 4;\]

\[\text{\ \ }x_{2} = 1 - 3 = - 2.\]

\[x \in ( - 2;3).\]

\[\textbf{б)}\ \left\{ \begin{matrix} 2x^{2} - 13x + 6 < 0 \\ x^{2} - 4x > 0\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 2 \cdot (x - 6)(x - 0,5) < 0 \\ x \cdot (x - 4) > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[2x^{2} - 13x + 6 = 0\]

\[D = 169 - 48 = 121\]

\[x_{1} = \frac{13 + 11}{4} = 6;\ \ \]

\[x_{2} = \frac{13 - 11}{4} = 0,5.\]

\[x \in (4;6).\]

\[\textbf{в)}\ \left\{ \begin{matrix} x^{2} - 6x - 16 > 0\ \ \\ x^{2} + 2x - 120 < 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x - 8)(x + 2) > 0\ \ \ \ \\ (x + 12)(x - 10) < 0 \\ \end{matrix} \right.\ \]

\[1)\ x^{2} - 6x - 16 = 0\]

\[D_{1} = 9 + 16 = 25\]

\[x_{1} = 3 + 5 = 8;\ \ x_{2} =\]

\[= 3 - 5 = - 2.\]

\[2)\ x^{2} + 2x - 120 = 0\]

\[D_{1} = 1 + 120 = 121\]

\[x_{1} = - 1 + 11 = 10;\ \ \ \]

\[x_{2} = - 1 - 11 = 12.\]

\[x \in ( - 12;\ - 2) \cup (8;10).\]

\[\textbf{г)}\ \left\{ \begin{matrix} 3x^{2} + x - 2 \leq 0\ \\ x^{2} + 4x - 12 \leq 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 3 \cdot \left( x - \frac{2}{3} \right)(x + 1) \leq 0 \\ (x + 6)(x - 2) \leq 0\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ 3x^{2} + x - 2 = 0\]

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{- 1 + 5}{6} = \frac{4}{6} = \frac{2}{3};\ \]

\[\ x_{2} = \frac{- 1 - 5}{6} = - 1.\]

\[2)\ x^{2} + 4x - 12 = 0\]

\[D_{1} = 4 + 12 = 16\]

\[x_{1} = - 2 + 4 = 2;\]

\[= \ \ x_{2} = - 2 - 4 = - 6.\]

\[x \in \left\lbrack - 1;\ \frac{2}{3} \right\rbrack.\]

\[\textbf{д)}\ \left\{ \begin{matrix} x^{2} + 4x + 15 > 0\ \ (1) \\ x^{2} - 9x + 8 \leq 0\ \ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ x^{2} + 4x + 15 = 0\ \]

\[D = 16 - 2 \cdot 4 \cdot 15 < 0 \Longrightarrow\]

\[\Longrightarrow x - любое\ число.\]

\[(2)\ x^{2} - 9x + 8 = 0\]

\[x_{1} + x_{2} = 9;\ \ x_{1} \cdot x_{2} = 8\]

\[x_{1} = 1;\ \ x_{2} = 8.\ \]

\[(x - 1)(x - 8) \leq 0\]

\[x \in \lbrack 1;8\rbrack.\]

\[\textbf{е)}\ \left\{ \begin{matrix} 2x^{2} + 5x - 3 < 0\ \ (1) \\ 3x^{2} + x + 11 < 0\ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ \ 2x^{2} + 5x - 3 = 0\ \]

\[D = 25 + 4 \cdot 2 \cdot 3 = 49\]

\[x_{1} = \frac{- 5 + 7}{4} = 0,5;\ \ \ \]

\[\ x_{2} = \frac{- 5 - 7}{4} = - 3;\]

\[(2)\ 3x^{2} + x + 11 = 0\]

\[D = 1 - 4 \cdot 3 \cdot 11 < 0 \Longrightarrow\]

\[\Longrightarrow решений\ нет.\]

\[Ответ:\ решений\ нет.\]