Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 592

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\[1)\ \frac{2x}{x + 1} + \frac{3x}{x - 1} = \frac{6x}{x^{2} - 1};\ \]

\[\text{\ \ \ x} \neq \pm 1\]

\[2x(x - 1) + 3x(x + 1) - 6x = 0\]

\[2x^{2} - 2x + 3x^{2} + 3x - 6x = 0\]

\[5x^{2} + x - 6x = 0\]

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

\[5x(x - 1) = 0\]

\[x = 0;\ \ \ x = 1\ (не\ подходит).\]

\[Ответ:x = 0.\]

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

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

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

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

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

\[Ответ:\ \ корней\ нет.\]

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

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

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

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

\[Ответ:\ \ x_{1} = 5\ \ и\ \ x_{2} = 6.\]

\[4)\ (x - 2)\left( x^{2} + 1 \right) = 2\left( x^{2} + 1 \right);\]

\[x - 2 = 2\]

\[x = 4.\]

\[Ответ:\ \ x = 4.\]