Решебник по алгебре 9 класс Макарычев Задание 298

 

\[\boxed{\text{298\ (298).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left( \frac{x + 2}{x - 4} \right)^{2} + 16 \cdot \left( \frac{x - 4}{x + 2} \right)^{2} = 17\]

\[Пусть\ t = \left( \frac{x + 2}{x - 4} \right)^{2}:\]

\[t + \frac{16}{t} - 17 = 0\ \ \ \ \ \ \ \ \ \ \ | \cdot t \neq 0\]

\[t^{2} - 17t + 16 = 0\]

\[t_{1} + t_{2} = 17;\ \ \ t_{1} \cdot t_{2} = 26\]

\[t_{1} = 1,\ \ t_{2} = 16;\]

\[1)\ при\ \ t_{1} = 1:\]

\[\left( \frac{x + 2}{x - 4} \right)^{2} = 1\ \]

\[\left\{ \begin{matrix} \frac{x + 2}{x - 4} = 1\ \ \\ \frac{x + 2}{x - 4} = - 1 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x + 2 = x - 4\ \ \ \\ x + 2 = - x + 4 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 2 = - 4 \\ 2x = 2 \\ \end{matrix} \right.\ \Longrightarrow x = 1.\]

\[2)\ при\ \ t_{2} = 16:\]

\[\left( \frac{x + 2}{x - 4} \right)^{2} = 16\]

\[\left\{ \begin{matrix} \frac{x + 2}{x - 4} = 4\ \ \ \ \\ \frac{x + 2}{x - 4} = - 4 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x + 2 = 4x - 16\ \ \ \\ x + 2 = - 4x + 16 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 3x = 18 \\ 5x = 14 \\ \end{matrix} \right.\ \Longrightarrow x_{1} =\]

\[= 6;\ \ x_{2} = 2,8.\]

\[Ответ:1;2,8;6.\]

\[\textbf{б)}\ \left( \frac{x + 1}{x - 3} \right)^{2} + 18 \cdot \left( \frac{x - 3}{x + 1} \right)^{2} = 11\]

\[Пусть\ \ t = \left( \frac{x + 1}{x - 3} \right)^{2}:\]

\[t + \frac{18}{t} = 11\ \ \ \ \ \ \ \ \ \ \ \ \ | \cdot t;\ \ t \neq 0\]

\[t^{2} - 11t + 18 = 0,\]

\[t_{1} = 2;\ \ \ \ \ t_{2} = 9.\]

\[1)\ при\ t_{1} = 2:\ \ \]

\[\left( \frac{x + 1}{x - 3} \right)^{2} = 2\]

\[\left\{ \begin{matrix} \frac{x + 1}{x - 3} = \sqrt{2\ \ } \\ \frac{x + 1}{x - 3} = - \sqrt{2} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x + 1 = \sqrt{2x} - 3\sqrt{2}\text{\ \ } \\ x + 1 = - \sqrt{2x} + 3\sqrt{2} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 7 + 4\sqrt{2} \\ x = 7 - 4\sqrt{2} \\ \end{matrix} \right.\ \]

\[2)\ при\ t_{2} = 9:\ \ \]

\[\left\{ \begin{matrix} \frac{x + 1}{x - 3} = 3 \\ \frac{x + 1}{x - 3} = - 3 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x + 1 = 3x - 9\ \ \ \ \\ x + 1 = - 3x + 9 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 5 \\ x = 2 \\ \end{matrix} \right.\ .\]

\[Ответ:\ \ 2;5;7 + 4\sqrt{2};7 - 4\sqrt{2}.\]

\[\boxed{\text{298.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{5x + 4}{x} < 4^{\backslash x}\]

\[\frac{5x + 4 - 4x}{x} < 0\]

\[\frac{x + 4}{x} < 0\]

\[x(x + 4) < 0\]

\[x \in ( - 4;0).\]

\[\textbf{б)}\ \frac{6x + 1}{x + 1} > 1^{\backslash x + 1}\]

\[\frac{6x + 1 - x - 1}{x + 1} > 0\]

\[\frac{5x}{x + 1} > 0\]

\[5x(x + 1) > 0\]

\[x \in ( - \infty;\ - 1) \cup (0; + \infty).\]

\[\textbf{в)}\ \frac{x}{x - 1} \geq 2^{\backslash x - 1}\]

\[\frac{x - 2x + 2}{x - 1} \geq 0\]

\[\frac{- x + 2}{x - 1} \geq 0\]

\[\frac{x - 2}{x - 1} \leq 0\]

\[(x - 2)(x - 1) \leq 0\]

\[x \in (1;2\rbrack.\]

\[\textbf{г)}\ \frac{3x - 1}{x + 2} \geq 1\]

\[\frac{3x - 1}{x + 2} - 1^{\backslash x + 2} \geq 0\]

\[\frac{3x - 1 - x - 2}{x + 2} \geq 0\]

\[\frac{2x - 3}{x + 2} \geq 0\]

\[(2x - 3)(x + 2) \geq 0\]

\[2 \cdot (x + 2)(x - 1,5) \geq 0\]

\[x \in ( - \infty;\ - 2) \cup \lbrack 1,5;\ + \infty).\]