Решебник по алгебре 8 класс Мерзляк ФГОС Задание 206

\[\boxed{\text{206\ (206).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 2x - 3 = 4\]

\[2x = 7\]

\[x = 3,5\]

\[Уравнение,\ равносильное\ \]

\[данному:\]

\[x - 3,5 = 0.\]

\[2)\ |x| = 1\]

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

\[Уравнение,\ равносильное\ \]

\[данному:\]

\[(x - 1)(x + 1) = 0.\]

\[3)\ x + 6 = x - 2\]

\[x - x = - 2 - 6\]

\[0 = - 8\]

\[Нет\ корней.\]

\[Уравнение,\ равносильное\ \]

\[данному:\]

\[x^{2} = - 5.\]

\[\boxed{\text{206.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\frac{x - 6}{x - 4} = 0\]

\[\left\{ \begin{matrix} x - 6 = 0\ \\ x - 4 \neq 0\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 6\ \\ x \neq 4\ \\ \end{matrix} \right.\ \]

\[Ответ:x = 6.\]

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

\[\left\{ \begin{matrix} x - 2 = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 2)(x + 2) \neq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x = 2\ \ \ \\ x \neq 2\ \ \ \\ x \neq - 2 \\ \end{matrix} \right.\ \]

\[Ответ:нет\ корней.\]

\[3)\ \frac{x^{2} - 4}{x - 2} = 0\]

\[\left\{ \begin{matrix} (x - 2)(x + 2) = 0 \\ x - 2 \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 2\ \ \ \\ x = - 2 \\ x \neq 2\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\ x = - 2.\]

\[4)\ \frac{x - 2}{x - 2} = 1\]

\[\left\{ \begin{matrix} x \neq 2 \\ 1 = 1 \\ \end{matrix} \right.\ \]

\[Ответ:x \in ( - \infty;2)\ \cup (2; + \infty).\]

\[5)\ \frac{2x^{2} + 18}{x^{2} + 9} = 2;\ \ \ x^{2} + 9 \neq 0\]

\[\frac{2 \cdot \left( x^{2} + 9 \right)}{x^{2} + 9} = 2\]

\[2 = 2\]

\[Ответ:x \in ( - \infty; + \infty).\]

\[6)\ \frac{x}{x - 5} + \frac{2x - 9}{x - 5} = 0\]

\[\frac{x + 2x - 9}{x - 5} = 0\]

\[\frac{3x - 9}{x - 5} = 0\]

\[\left\{ \begin{matrix} 3x - 9 = 0 \\ x - 5 \neq 0\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x = 3 \\ x \neq 5 \\ \end{matrix} \right.\ \]

\[Ответ:x = 3.\]

\[7)\ \frac{5x - 7}{x + 1} - \frac{x - 5}{x + 1} = 0\]

\[\frac{5x - 7 - x + 5}{x + 1} = 0\]

\[\frac{4x - 2}{x + 1} = 0\]

\[\left\{ \begin{matrix} 4x - 2 = 0 \\ x + 1 \neq 0\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 0,5 \\ x \neq - 1 \\ \end{matrix} \right.\ \]

\[Ответ:x = 0,5.\]

\[8)\ \frac{2x + 16}{x + 3} - \frac{1 - 3x}{x + 3} = 0\]

\[\frac{2x + 16 - 1 + 3x}{x + 3} = 0\]

\[\frac{5x + 15}{x + 3} = 0\]

\[\left\{ \begin{matrix} 5x + 15 = 0 \\ x + 3 \neq 0\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - 3 \\ x \neq - 3 \\ \end{matrix} \right.\ \]

\[Ответ:нет\ корней.\]

\[9)\ \frac{2^{\backslash x + 1}}{x - 1} + \frac{1^{\backslash x - 1}}{x + 1} = 0\]

\[\frac{2x + 2 + x - 1}{(x - 1)(x + 1)} = 0\]

\[\frac{3x + 1}{(x - 1)(x + 1)} = 0\]

\[\left\{ \begin{matrix} 3x + 1 = 0 \\ x \neq 1\ \ \ \ \ \ \ \ \ \ \\ x \neq - 1\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - \frac{1}{3} \\ x \neq 1\ \ \ \ \\ x \neq - 1 \\ \end{matrix} \right.\ \]

\[Ответ:\ x = - \frac{1}{3}.\]

\[10)\ \frac{3}{x - 2} = \frac{4}{x + 3}\]

\[\frac{3^{\backslash x + 3}}{x - 2} - \frac{4^{\backslash x - 2}}{x + 3} = 0\]

\[\frac{3x + 9 - 4x + 8}{(x - 2)(x + 3)} = 0\]

\[\left\{ \begin{matrix} - x + 17 = 0 \\ x \neq 2\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq - 3\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 17 \\ x \neq 2\ \ \ \\ x \neq - 3 \\ \end{matrix} \right.\ \]

\[Ответ:x = 17.\]

\[11)\ \frac{x}{x - 6} = 2\]

\[\frac{x}{x - 6} - 2^{\backslash x - 6} = 0\]

\[\frac{x - 2x + 12}{x - 6} = 0\]

\[\left\{ \begin{matrix} - x + 12 = 0 \\ x - 6 \neq 0\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 12 \\ x \neq 6\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x = 12.\]

\[12)\ \frac{x - 4^{\backslash 2x - 1}}{x - 3} - \frac{2x + 1^{\backslash x - 3}}{2x - 1} = 0\]

\[\frac{2x^{2} - 8x - x + 4 - 2x^{2} - x + 6x + 3}{(x - 3)(2x - 1)} = 0\]

\[\frac{- 4x + 7}{(x - 3)(2x - 1)} = 0\]

\[\left\{ \begin{matrix} - 4x + 7 = 0 \\ x \neq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq 0,5\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 1\frac{3}{4} \\ x \neq 3\ \ \ \ \\ x \neq 0,5 \\ \end{matrix} \right.\ \]

\[Ответ:x = 1\frac{3}{4}.\]

\[13)\ \frac{x + 8^{\backslash x - 2}}{x} - \frac{6^{\backslash x}}{x - 2} = 0\]

\[\frac{x^{2} + 8x - 2x - 16 - 6x}{x(x - 2)} = 0\]

\[\left\{ \begin{matrix} x^{2} - 16 = 0 \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq 2\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 4\ \ \ \\ x = - 4 \\ x \neq 0\ \ \ \\ x \neq 2\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x = 4;\ x = - 4.\]

\[14)\ \frac{2x^{\backslash x + 5}}{x - 5} - \frac{x^{2} + 15x}{x^{2} - 25} = 0\]

\[\frac{2x^{2} + 10x - x^{2} - 15x}{(x - 5)(x + 5)} = 0\]

\[\left\{ \begin{matrix} x(x - 5) = 0 \\ x \neq 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq - 5\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 0\ \ \ \\ x = 5\ \ \ \\ x \neq 5\ \ \ \\ x \neq - 5 \\ \end{matrix} \right.\ \]

\[Ответ:\ \ x = 0.\]