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МатематикаРешебник по алгебре 9 класс Макарычев Задание 394
Задание 394
\[\boxed{\text{394\ (394).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{6x + 2}{x + 4} < 5^{\backslash x + 4}\]
\[\frac{6x + 2 - 5 \cdot (x + 4)}{x + 4} < 0\]
\[\frac{6x + 2 - 5x - 20}{x + 4} < 0\]
\[\frac{x - 18}{x + 4} < 0\]
\[(x + 4)(x - 18) < 0\]
\[x \in ( - 4;\ 18).\]
\[\textbf{б)}\ \frac{5x + 8}{x} > 1^{\backslash x}\]
\[\frac{5x + 8 - x}{x} > 0\]
\[\frac{4x + 8}{x} > 0\]
\[x(4x + 8) > 0\]
\[4x(x + 2) > 0\]
\[x \in ( - \infty; - 2) \cup (0; + \infty).\]
\[\textbf{в)}\ \frac{3 - 2x}{3x + 2} \leq 1^{\backslash 3x + 2}\]
\[\frac{3 - 2x - 3x - 2}{3x + 2} \leq 0\]
\[\frac{1 - 5x}{3x + 2} \leq 0\]
\[\frac{5x - 1}{3x + 2} \geq 0\]
\[\left\{ \begin{matrix} (5x - 1)(3x + 2) \geq 0 \\ x \neq - \frac{2}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 15 \cdot \left( x + \frac{2}{3} \right)(x - 0,2) \geq 0 \\ x \neq - \frac{2}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[x \in \left( - \infty;\ - \frac{2}{3} \right) \cup \lbrack 0,2; + \infty).\]
\[\textbf{г)}\ \frac{5x - 4}{x + 8} \geq 15^{\backslash x + 8}\]
\[\frac{5x - 4 - 15 \cdot (x + 8)}{x + 8} \geq 0\]
\[\frac{- 10x - 124}{x + 8} \geq 0\]
\[\frac{x + 12,4}{x + 8} \geq 0\]
\[\left\{ \begin{matrix} (x + 12,4)(x + 8) \geq 0 \\ x \neq - 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x \in \lbrack - 12,4;\ - 8).\]
\[\boxed{\text{394.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 36 \\ y = x^{2} + 6\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x^{2} + y^{2} = 36 \\ x^{2} = y - 6\ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y^{2} + y - 42 = 0 \\ x^{2} = y - 6\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[y^{2} + y - 42 = 0\]
\[y_{1} + y_{2} = - 1;\ \ \ y_{1} \cdot y_{2} = - 42\]
\[y_{1} = 6;\ \ \ y_{2} = - 7.\]
\[\Longrightarrow \left\{ \begin{matrix} y = 6 \\ x = 0 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \ \]
\[\left\{ \begin{matrix} y = - 7\ \ \ \ \\ x^{2} = - 13 \\ \end{matrix} \right.\ \Longrightarrow корней\ нет.\]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 16\ \ \ \ \ \ \ \ \ \ \\ (x - 2)^{2} + y^{2} = 36 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y^{2} = 16 - x^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ (x - 2)^{2} + 16 - x^{2} = 36 \\ \end{matrix} \right.\ \]
\[(x - 2)^{2} + 16 - x^{2} = 36\]
\[x^{2} - 4x + 4 + 16 - x^{2} = 36\]
\[- 4x = 36 - 20\]
\[- 4x = 16\]
\[x = - 4.\]
\[\left\{ \begin{matrix} y^{2} = 16 - 4^{2} \\ x = - 4\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y^{2} = 0 \\ x = - 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 0 \\ x = - 4 \\ \end{matrix}. \right.\ \]
\[\textbf{в)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 25 \\ 4x - y = 0\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} y^{2} = 25 - x^{2} \\ y = 4x\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = 4x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (4x)^{2} = 25 - x^{2} \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y = 4x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 16x^{2} + x^{2} = 25 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} y = 4x\ \ \ \ \ \ \\ 17x^{2} = 25 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} x = \pm \sqrt{\frac{25}{17}} \\ y = 4x\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x = \pm 1,25 \\ y = \pm 5\ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (0;6);\ \ б)\ ( - 4;0);\ \ \]
\[\textbf{в)}\ ( - 1,25; - 5);(1,25;5)\text{.\ }\]
