Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 142

\[\boxed{\mathbf{142.}}\]

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

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

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

\[\frac{3x^{2} - 3x}{x^{2} - 1} = 0\]

\[\frac{3x(x - 1)}{(x - 1)(x + 1)} = 0\]

\[\frac{3x}{x + 1} = 0;\ \ \ \ x \neq - 1\]

\[3x = 0\]

\[x = 0.\]

\[Ответ:\ \ x = 0.\]

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

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

\[\frac{x^{2} - x - 2x + 4}{x(x - 2)} - \frac{x}{x(x - 2)} = 0\]

\[\frac{x^{2} - 4x + 4}{x(x - 2)} = 0\]

\[\frac{(x - 2)^{2}}{x(x - 2)} = 0;\ \ \ \ \ x \neq 2;\ \ \ \ x \neq 0\]

\[x = 2.\]

\[Ответ:\ \ корней\ нет.\]

\[3)\ (x - 3)(x - 5) = 3(x - 5)\]

\[(x - 3)(x - 5) - 3(x - 5) = 0\]

\[(x - 5)(x - 3 - 3) = 0\]

\[(x - 5)(x - 6) = 0\]

\[x = 5;\ \ \ x = 6.\]

\[Ответ:\ \ x = 5;\ \ x = 6.\]

\[x - 2 = 2\]

\[\ x = 4.\]

\[Ответ:\ \ x = 4.\]