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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 385
Задание 385
\[\boxed{\text{385\ (385).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + x - 6 < 0\ \ \ \\ - x^{2} + 2x + 3 > 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x^{2} + x - 6 < 0\ \ \ \ (1) \\ x^{2} - 2x - 3 < 0\ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\ x^{2} + x - 6 = 0\ \ \]
\[D = 1 + 4 \cdot 6 = 25\ \ \]
\[x_{1,2} = \frac{- 1 \pm 5}{2} = - 3;2.\]
\[(2)\text{\ \ }x^{2} - 2x - 3 = 0\ \ \]
\[D_{1} = 1 + 3 = 4\]
\[x_{1,2} = 1 \pm 2 = - 1;3.\]
\[\left\{ \begin{matrix} (x + 3)(x - 2) < 0 \\ (x + 1)(x - 3) < 0 \\ \end{matrix} \right.\ \]
\[x \in ( - 1;2).\]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + 4x - 5 > 0\ \ (1) \\ x^{2} - 2x - 8 < 0\ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\ x^{2} + 4x - 5 = 0\ \ \]
\[D_{1} = 2^{2} + 5 = 9\]
\[x_{1,2} = - 2 \pm 3 = - 5;1.\]
\[(2)\ x^{2} - 2x - 8 = 0\ \ \]
\[D_{1} = 1 + 8 = 9\]
\[x_{1,2} = 1 \pm 3 = - 2;4.\]
\[\left\{ \begin{matrix} (x + 5)(x - 1) > 0 \\ (x + 2)(x - 4) < 0 \\ \end{matrix} \right.\ \]
\[x \in (1;4).\]
\[\boxed{\text{385.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 16 \\ x - y = 4\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (y + 4)^{2} + y^{2} = 16 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 4\ \ \ \ \ \\ y(y + 4) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 0 \\ x_{1} = 4 \\ \end{matrix} \right.\ \text{\ \ }или\ \ \left\{ \begin{matrix} y_{2} = - 4 \\ x_{2} = 0\ \ \ . \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} + y^{2} = 16 \\ x - y = 4\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x^{2} + y^{2} = 16 \\ y = x - 4\ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} y = x^{2} + 1 \\ x + 2y = 5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = x^{2} + 1\ \ \ \ \ \ \ \ \ \\ 2x^{2} + x - 3 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 1 \\ y_{1} = 2 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \left\{ \begin{matrix} x_{2} = - 1,5 \\ y_{2} = 3,25. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (4;0);(0;\ - 4);\ \ \ \]
\[\textbf{б)}\ (1;2);\ \ ( - 1,5;3,25).\]
