Решите уравнение 3/(x^2+4x+4)+4/(x^2-4)=1/(x-2).

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

\[\frac{3}{(x + 2)^{2}} + \frac{4}{(x - 2)(x + 2)} - \frac{1}{x - 2} = 0\]

\[ОДЗ:\ \ x \neq 2;\ \ x \neq - 2.\]

\[3 \cdot (x - 2) + 4 \cdot (x + 2) - (x + 2)^{2} = 0\]

\[3x - 6 + 4x + 8 - x^{2} - 4x - 4 = 0\]

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

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

\[x_{1} + x_{2} = 3;\ \ \ x_{1} \cdot x_{2} = 2\]

\[x_{1} = 2\ (не\ подходит);\ \ x_{2} = 1.\]

\[Ответ:x = 1.\]