ГДЗ по алгебре 8 класс Мерзляк Задание 218

\[\boxed{\text{218}\text{\ (218)}\text{.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

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

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

\[Ответ:y = 0,6.\]

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

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

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

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

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

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

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

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

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

\[нет.\]

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

\[Ответ:y = 0.\]