ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 598

\[\boxed{\mathbf{598}.}\]

\[1)\ \frac{3}{x - 1} - \frac{4x - 1}{x + 1} = \frac{x^{2} + 5}{x^{2} - 1} - 5;\]

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

\[= \frac{x^{2} + 5 - 5\left( x^{2} - 1 \right)}{x^{2} - 1};\]

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

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

\[\frac{- 4x^{2} + 8x + 2}{x^{2} - 1} = \frac{- 4x^{2} + 10}{x^{2} - 1};\]

\[\frac{8x - 8}{x^{2} - 1} = 0;\]

\[\frac{8(x - 1)}{(x - 1)(x + 1)} = 0;\]

\[\frac{8}{x + 1} = 0;\]

\[Ответ:\ \ корней\ нет.\]

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

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

\[\frac{x + 2}{x - 2} - \frac{x - 2}{x + 2} =\]

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

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

\[= \frac{x^{2} - 4x + 12 + 4x}{x^{2} - 4};\]

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

\[= \frac{x^{2} + 12}{x^{2} - 4};\]

\[\frac{x^{2} - 8x + 12}{x^{2} - 4} = 0;\]

\[x^{2} - 8x + 12 = 0;\]

\[D = 8^{2} - 4 \bullet 12 = 64 - 48 = 16\]

\[x_{1} = \frac{8 - 4}{2} = 2\ \ и\]

\[\text{\ \ }x_{2} = \frac{8 + 4}{2} = 6;\]

\[Выражение\ имеет\ смысл\ при:\]

\[x^{2} - 4 \neq 0;\]

\[x^{2} \neq 4;\]

\[x \neq \pm 2;\]

\[Ответ:\ \ x = 6.\]