ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 805

\[\boxed{\text{805.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\frac{6}{y + 1} + \frac{y}{y - 2} = \frac{6}{y + 1} \cdot \frac{y}{y - 2}\]

\[\frac{6y - 12 + y^{2} + y}{(y + 1)(y - 2)} =\]

\[= \frac{6y}{(y + 1)(y - 2)},\]

\[y + 1 \neq 0,\ \ y \neq - 1\]

\[y - 2 \neq 0,\ \ y \neq 2\]

\[y^{2} + 7y - 12 = 6y\]

\[y^{2} + y - 12 = 0\]

\[D = 1 + 48 = 49\]

\[y_{1,2} = \frac{- 1 \pm 7}{2} = 3;\ - 4\]

\[Ответ:y = \left\{ - 4;3 \right\}.\]

\[\textbf{б)}\frac{2}{y - 3} + \frac{6}{y + 3} =\]

\[= \frac{2}{y - 3}\ :\frac{6}{y + 3}\]

\[\frac{2y + 6 + 6y - 18}{(y - 3)(y + 3)} =\]

\[= \frac{2}{y - 3} \cdot \frac{y + 3}{6}\]

\[\frac{8y - 12}{y^{2} - 9} = \frac{y + 3}{(y - 3) \cdot 3}\]

\[y^{2} - 9 \neq 0,\ \ y \neq \pm 3\]

\[3 \cdot (8y - 12)(y - 3) =\]

\[= (y + 3)\left( y^{2} - 9 \right)\]

\[3 \cdot (8y - 12)(y - 3) =\]

\[= (y + 3)(y + 3)(y - 3)\]

\[24y - 36 = y^{2} + 6y + 9\]

\[y^{2} - 18y + 45 = 0\]

\[D = 324 - 180 = 144\]

\[y_{1,2} = \frac{18 \pm 12}{2} =\]

\[= 15;3 - не\ подходит\ по\ ОДЗ\]

\[Ответ:y = 15.\]

\[\textbf{в)}\frac{y + 12}{y - 4} - \frac{y}{y + 4} =\]

\[= \frac{y + 12}{y - 4} \cdot \frac{y}{y + 4}\]

\[y \neq \pm 4\]

\[(y + 12)(y + 4) -\]

\[- y(y - 4) = y(y + 12)\]

\[y^{2} + 4y + 12y + 48 -\]

\[- y^{2} + 4y = y^{2} + 12y\]

\[y^{2} - 8y - 48 = 0\]

\[D = 64 + 192 = 256\]

\[y_{1,2} = \frac{8 \pm 16}{2} = 12;\ \]

\[- 4 - не\ подходит\ по\ ОДЗ\]

\[Ответ:y = 12\text{.\ }\]