Answer
$ \frac{-x+14}{7} ; x \ne +2,-2;$
Work Step by Step
$\frac{\frac{3}{x-2} - \frac{4}{x+2}}{\frac{7}{x^{2}-4}}$
Taking Numerator,
$\frac{3}{x-2} - \frac{4}{x+2} = \frac{3(x+2)-4(x-2)}{(x-2)(x+2)} ; x \ne +2,-2;$
$= \frac{3x+6-4x+8}{(x-2)(x+2)}; x \ne +2,-2;$
$= \frac{-x+14}{(x-2)(x+2)}; x \ne +2,-2;$
Taking the Denominator,
$\frac{7}{x^{2}-4} = \frac{7}{(x-2)(x+2)}; x \ne +2,-2;$
$[(a^{2}-b^{2}) = (a+b)(a-b)]$
The given expression becomes,
$\frac{\frac{3}{x-2} - \frac{4}{x+2}}{\frac{7}{x^{2}-4}} = $$\frac{\frac{-x+14}{(x-2)(x+2)}}{\frac{7}{(x-2)(x+2)}}; x \ne +2,-2;$
$= \frac{-x+14}{(x-2)(x+2)} \times \frac{(x-2)(x+2)}{7}; x \ne +2,-2;$
Divide out common factors.
$= \frac{-x+14}{7} ; x \ne +2,-2;$
