ГДЗ по алгебре 8 класс Мерзляк Задание 212

\[\boxed{\text{212\ (212).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{5}{x^{2} - 4} + \frac{2x^{\backslash x - 2}}{x + 2} = 2^{\backslash x^{2} - 4}\]

\[\frac{5 + 2x(x - 2) - 2 \cdot \left( x^{2} - 4 \right)}{(x - 2)(x + 2)} = 0\]

\[\frac{5 + 2x^{2} - 4x - 2x^{2} + 8}{(x - 2)(x + 2)} = 0\]

\[\frac{- 4x + 13}{(x - 2)(x + 2)} = 0\]

\[\left\{ \begin{matrix} - 4x + 13 = 0 \\ x \neq 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq - 2\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = \frac{13}{4}\ \\ x \neq 2\ \ \ \\ x \neq - 2 \\ \end{matrix} \right.\ \]

\[Ответ:x = 3\frac{1}{4}.\]

\[2)\ \frac{2^{\backslash 6x - 1}}{6x + 1} + \frac{3^{\backslash 6x + 1}}{6x - 1} = \frac{30x + 9}{36x^{2} - 1}\]

\[\frac{2 \cdot (6x - 1) + 3 \cdot (6x + 1) - 30x - 9}{(6x + 1)(6x - 1)} = 0\]

\[\frac{12x - 2 + 18x + 3 - 30x - 9}{(6x + 1)(6x - 1)} = 0\]

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

\[- 8 \neq 0\]

\[Ответ:нет\ корней.\]

\[3)\ \frac{6x + 14}{x^{2} - 9} + \frac{7}{x^{2} + 3x} = \frac{6}{x - 3}\]

\[\frac{6x + 14^{\backslash x}}{(x - 3)(x + 3)} + \frac{7^{\backslash x - 3}}{x(x + 3)} =\]

\[= \frac{6^{\backslash x(x + 3)}}{x - 3}\]

\[\frac{x(6x + 14) + 7 \cdot (x - 3) - 6 \cdot x(x + 3)}{x(x - 3)(x + 3)} = 0\]

\[\frac{6x^{2} + 14x + 7x - 21 - 6x^{2} - 18}{x(x - 3)(x + 3)} = 0\]

\[\frac{3x - 21}{x(x - 3)(x + 3)} = 0\]

\[\left\{ \begin{matrix} 3x - 21 = 0 \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq 3\ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq - 3\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x = 7\ \ \ \\ x \neq 0\ \ \ \\ x \neq 3\ \ \ \\ x \neq - 3 \\ \end{matrix} \right.\ \]

\[Ответ:x = 7.\]

\[4)\ \frac{2y^{2} + 5}{1 - y^{2}} + \frac{y + 1}{y - 1} = \frac{4}{y + 1}\]

\[\frac{2y^{2} + 5}{(1 - y)(1 + y)} - \frac{y + 1^{\backslash 1 + y}}{1 - y} -\]

\[- \frac{4^{\backslash 1 - y}}{1 + y} = 0\]

\[\frac{2y^{2} + 5 - (1 + y)(1 + y) - 4 \cdot (1 - y)}{(1 - y)(1 + y)} = 0\]

\[\frac{2y^{2} + 5 - 1 - 2y - y^{2} - 4 + 4y}{(1 - y)(1 + y)} = 0\]

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

\[\left\{ \begin{matrix} y(y + 2) = 0 \\ y \neq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ y \neq - 1\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 0\ \ \ \\ y = - 2 \\ y \neq 1\ \ \ \\ y \neq - 1 \\ \end{matrix} \right.\ \]

\[Ответ:y = 0;\ y = - 2.\]

\[5)\ \frac{2x - 1}{2x + 1} = \frac{2x + 1}{2x - 1} + \frac{4}{1 - 4x^{2}}\]

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

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

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

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

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

\[\left\{ \begin{matrix} - 8x + 4 = 0 \\ x \neq 0,5\ \ \ \ \ \ \ \ \ \\ x \neq - 0,5\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 0,5\ \ \\ x \neq 0,5\ \ \\ x \neq - 0,5 \\ \end{matrix} \right.\ \]

\[Ответ:нет\ корней.\]

\[6)\ \frac{7^{\text{(}x + 2)(x - 3)}}{(x + 2)(x - 3)} - \frac{4^{{\backslash(x + 2)}^{2}}}{(x - 3)^{2}} =\]

\[= \frac{3^{{\backslash(x - 3)}^{2}}}{(x + 2)^{2}}\]

\[\frac{- 5x - 85}{(x + 2)^{2}(x - 3)^{2}} = 0\]

\[\left\{ \begin{matrix} - 5x - 85 = 0 \\ x \neq - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - 17 \\ x \neq - 2\ \ \\ x \neq 3\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\ x = - 17.\ \]

\[7)\ \frac{2x - 1}{x + 4} - \frac{3x - 1}{4 - x} = \frac{6x + 64}{x^{2} - 16} + 4\]

\[\frac{2x - 1^{\backslash x - 4}}{x + 4} + \frac{3x - 1^{\backslash x + 4}}{x - 4} -\]

\[- \frac{6x + 64}{(x - 4)(x + 4)} - 4^{\backslash x^{2} - 16} = 0\]

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

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

\[Ответ:x = 0.\ \]

\[8)\ \frac{2x - 6}{x^{2} - 36} - \frac{x - 3}{x^{2} - 6x} -\]

\[- \frac{x - 1}{x^{2} + 6x} = 0\]

\[\frac{2x - 6^{\backslash x}}{(x - 6)(x + 6)} - \frac{x - 3^{\backslash x + 6}}{x(x - 6)} -\]

\[- \frac{x - 1^{\backslash x - 6}}{x(x + 6)} = 0\]

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

\[\left\{ \begin{matrix} - 2x + 12 = 0 \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 6\ \ \ \\ x \neq 0\ \ \ \\ x \neq 6\ \ \ \\ x \neq - 6 \\ \end{matrix} \right.\ \]

\[Ответ:нет\ корней.\]