Решите уравнение 5/(x^2-4x+4)-4/(x^2-4)=1/(x+2).

\[\frac{5}{x^{2} - 4x + 4} - \frac{4}{x^{2} - 4} = \frac{1}{x + 2}\]

\[\frac{5}{(x - 2)^{2}} - \frac{4}{(x - 2)(x + 2)} - \frac{1}{x + 2} = 0\]

\[ОДЗ:\ \ x \neq 2;\ \ x \neq - 2.\]

\[5 \cdot (x + 2) - 4 \cdot (x - 2) - (x - 2)^{2} = 0\]

\[5x + 10 - 4x + 8 - x^{2} + 4x - 4 = 0\]

\[- x^{2} + 5x + 14 = 0\]

\[x^{2} - 5x - 14 = 0\]

\[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = - 14\]

\[x_{1} = 7;\ \ \ \ x_{2} = - 2.\]

\[Ответ:x = 7.\]