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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 388
Задание 388
\[\boxed{\text{388\ (388).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ (18x - 36)(x - 7) > 0\]
\[18 \cdot (x - 2)(x - 7) > 0\]
\[x \in ( - \infty;2) \cup (7; + \infty).\]
\[\textbf{б)}\ (x - 7,3)(9,8 - x) > 0\]
\[(x - 7,3)(x - 9,8) < 0\]
\[x \in (7,3;9,8).\]
\[\textbf{в)}\ (x + 0,8)(4 - x)(x - 20) < 0\]
\[(x + 0,8)(x - 4)(x - 20) > 0\]
\[x \in ( - 0,8;4) \cup (20; + \infty).\]
\[\textbf{г)}\ (10x + 3)(17 - x)(x - 5) \geq 0\]
\[10 \cdot (x + 0,3)(x - 5)(x - 17) \leq 0\]
\[x \in ( - \infty; - 0,3\rbrack \cup \lbrack 5;17\rbrack.\ \]
\[\boxed{\text{388.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} - 4 = 0 \\ xy = 6\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x^{2} = 4 \\ xy = 6 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = \pm 2 \\ y = \frac{6}{x}\text{\ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x_{1} = 2 \\ y_{1} = 3 \\ \end{matrix} \right.\ \text{\ \ }или\ \]
\[\ \left\{ \begin{matrix} x_{2} = - 2 \\ y_{2} = - 3 \\ \end{matrix} \right.\ .\]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} - 5x + 6 = 0 \\ y^{2} - 6y + 5 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x - 2)(x - 3) = 0 \\ (y - 1)(y - 5) = 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 5x + 6 = 0\]
\[x_{1} + x_{2} = 5;\ \ \ \ x_{1} \cdot x_{2} = 6\]
\[x_{1} = 2;\ \ \ \ \ x_{2} = 3.\]
\[y^{2} - 6y + 5 = 0\]
\[D_{1} = 9 - 5 = 4\]
\[y_{1} = 3 + 2 = 5;\ \ \]
\[\ y_{2} = 3 - 2 = 1.\]
\[\left\{ \begin{matrix} x_{1} = 2 \\ y_{1} = 1 \\ \end{matrix} \right.\ ;\ \ \ \ \left\{ \begin{matrix} x_{2} = 3 \\ y_{2} = 1 \\ \end{matrix} \right.\ ;\ \ \ \ \]
\[\left\{ \begin{matrix} x_{3} = 2 \\ y_{3} = 5 \\ \end{matrix} \right.\ ;\ \ \ \ \left\{ \begin{matrix} x_{4} = 3 \\ y_{4} = 5 \\ \end{matrix} \right.\ .\]
\[Ответ:а)\ (2;3);( - 2;\ - 3);\ \]
\[\ б)\ (2;1);(3;1);(2;5);(3;5).\]
