ГДЗ по алгебре 9 класс Макарычев Задание 494

\[\boxed{\text{494\ (494).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

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

\[x^{2} + 3x + 2 = (x + 2)(x + 1)\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = 2\]

\[x_{1} = - 2;\ \ \ x_{2} = - 1.\]

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

\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 2\]

\[x_{1} = - 2;\ \ x_{2} = 1.\]

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