Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 802

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

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

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

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

\[x^{2} - x - 2 = 0\]

\[D = 1 + 8 = 9\]

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

\[x_{2} = \frac{1 + 3}{2} = 2.\]

\[Ответ:\ \ 2.\]

\[2)\ \frac{2x^{2}}{x - 1} - \frac{3x}{x + 2} = \frac{2(4x - 1)}{x^{2} + x - 2}\]

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

\[= 2(4x - 1)\]

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

\[2x^{3} + x^{2} - 5x + 2 = 0\]

\[(x - 1)\left( 2x^{2} + 3x - 2 \right) = 0\]

\[D = 9 + 16 = 25\]

\[x_{1} = \frac{- 3 - 5}{2 \bullet 2} = - 2;\]

\[x_{2} = \frac{- 3 + 5}{2 \bullet 2} = \frac{1}{2}.\]

\[Ответ:\ \ \frac{1}{2}.\]