Решебник по алгебре 8 класс Мерзляк ФГОС Задание 217

 

\[\boxed{\text{217\ (217).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{x + 5}{x^{2} - 5x} - \frac{x - 5}{2x^{2} + 10x} =\]

\[= \frac{x + 25}{2x^{2} - 50}\]

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

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

\[\frac{2 \cdot (x + 5)^{2} - (x - 5)^{2} - x^{2} - 25x}{2x(x - 5)(x + 5)} = 0\]

\[\left\{ \begin{matrix} 5x + 25 = 0 \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq 5\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq - 5\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - 5 \\ x \neq 0\ \ \ \\ x \neq 5\ \ \ \\ x \neq - 5 \\ \end{matrix} \right.\ \]

\[Ответ:нет\ корней.\]

\[2)\ \frac{2}{x^{2} - 9} - \frac{1}{2x^{2} - 12x + 18} =\]

\[= \frac{3}{2x^{2} + 6x}\]

\[\frac{2^{\backslash 2x(x - 3)}}{(x - 3)(x + 3)} - \frac{1^{\backslash x(x + 3)}}{2 \cdot (x - 3)^{2}} -\]

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

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

\[\frac{3x - 27}{2x(x - 3)^{2}(x + 3)} = 0\]

\[\left\{ \begin{matrix} 3x - 27 = 0 \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq - 3\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 9\ \ \ \ \\ x \neq 0\ \ \ \ \\ x \neq 3\ \ \ \ \\ x \neq - 3 \\ \end{matrix} \right.\ \]

\[Ответ:x = 9.\]

\[3)\ \frac{9x + 12}{x^{3} - 64} - \frac{1}{x - 4} =\]

\[= \frac{1}{x^{2} + 4x + 16}\]

\[\frac{9x + 12}{(x - 4)\left( x^{2} + 4x + 16 \right)} -\]

\[- \frac{1^{\backslash x^{2} + 4x + 16}}{x - 4} - \frac{1^{\backslash x - 4}}{x^{2} + 4x + 16} = 0\]

\[\frac{9x + 12 - x^{2} - 4x - 16 - x + 4}{(x - 4)\left( x^{2} + 4x + 16 \right)} = 0\]

\[\frac{- x^{2} + 4x}{(x - 4)\left( x^{2} + 4x + 16 \right)} = 0\]

\[x^{2} + 4x + 16 \neq 0\]

\[D_{1} = 4 - 16 = - 12 < 0 - нет\]

\[\ корней.\]

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

\[Ответ:x = 0.\]

\[\boxed{\text{217.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{4y + 24}{5y^{2} - 45} + \frac{y + 3}{5y^{2} - 15y} =\]

\[= \frac{y - 3}{y^{2} + 3y}\]

\[\frac{4y + 24^{\backslash y}}{5 \cdot (y - 3)(y + 3)} + \frac{y + 3^{\backslash y + 3}}{5y(y - 3)} -\]

\[- \frac{y - 3^{\backslash 5(y - 3)}}{y(y + 3)} = 0\]

\[\frac{60y - 36}{5y(y - 3)(y + 3)} = 0\]

\[\left\{ \begin{matrix} 60y - 36 = 0 \\ y \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y \neq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y \neq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = \frac{6}{10}\text{\ \ } \\ y \neq 0\ \ \ \ \ \\ y \neq 3\ \ \ \ \ \\ y \neq - 3\ \\ \end{matrix} \right.\ \]

\[Ответ:y = 0,6.\]

\[2)\ \frac{y + 2}{8y^{3} + 1} - \frac{1}{4y + 2} =\]

\[= \frac{y + 3}{8y^{2} - 4y + 2}\]

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

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

\[- \frac{y + 3^{\backslash 2y + 1}}{2 \cdot \left( 4y^{2} - 2y + 1 \right)} = 0\]

\[\frac{- 6y^{2} - 3y}{2 \cdot (2y + 1)\left( 4y^{2} - 2y + 1 \right)} = 0\]

\[\frac{- 3y(2y + 1)}{2 \cdot (2y + 1)\left( 4y^{2} - 2y + 1 \right)} = 0\]

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

\[D_{1} = 1 - 4 = - 3 < 0 - корней\ \]

\[нет.\]

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

\[Ответ:y = 0.\]