Решите уравнение: 5/(x^2+2x+1)=2/(1-x^2 )+1/(x-1).

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

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

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

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

\[\frac{5}{(x + 1)^{2}\ } = \frac{1}{1 + x}\]

\[\frac{5}{(x + 1)^{2}} - \frac{1^{\backslash x + 1}}{x + 1} = 0\]

\[\frac{5 - (x + 1)}{(x + 1)^{2}} = 0\]

\[5 - x - 1 = 0\]

\[x = 4\]

\[Ответ:4.\]