Докажите тождество: (1/(x+1)+1/(x-1)+1/(x+2)+1/(x-2)-2x/(x^2-4))*(1/x+1/x^2)=2/(x^2-x).

Ответ:

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

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

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

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

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

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

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

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

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

\[Что\ и\ требовалось\ доказать.\]