Докажите тождество ((2x+5)/(x^2+4x+4)-(x+3)/(x^2+2x)) :(x^2-6)/(x^3-4x)=(x-2)/(x+2).

\[\left( \frac{2x + 5}{x^{2} + 4x + 4} - \frac{x + 3}{x^{2} + 2x} \right)\ :\frac{x^{2} - 6}{x^{3} - 4x} = \frac{x - 2}{x + 2}\]

\[Упростим\ левую\ часть\ тождества:\]

\[\left( \frac{2x + 5}{x^{2} + 4x + 4} - \frac{x + 3}{x^{2} + 2x} \right)\ :\frac{x^{2} - 6}{x^{3} - 4x} =\]

\[= \left( \frac{2x + 5^{\backslash x}}{(x + 2)^{2}} - \frac{x + 3^{\backslash x + 2}}{x(x + 2)} \right) \cdot \frac{x^{3} - 4x}{x^{2} - 6} =\]

\[= \frac{2x^{2} + 5x - x^{2} - 3x - 2x - 6}{x(x + 2)^{2}} \cdot \frac{x^{3} - 4x}{x^{2} - 6} =\]

\[= \frac{\left( x^{2} - 6 \right) \cdot x\left( x^{2} - 4 \right)}{x(x + 2)^{2} \cdot \left( x^{2} - 6 \right)} = \frac{(x - 2)(x + 2)}{(x + 2)^{2}} =\]

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