Answer
$2$; $x\ne-2, 2$
Work Step by Step
Before we multiply two fractions, we should factor each to see if we can cancel any terms.
Top-left: $x^{2}-4$ is a difference of squares and factors to $(x + 2)(x - 2)$.
Bottom-left: We have a perfect square trinomial, which factors to $(x - 2)(x - 2)$.
Top-right: We can pull out a greatest common factor of $2$, leaving us with $2(x - 2)$.
$\frac{(x+2)(x-2)}{(x-2)(x-2)}\times \frac{2(x-2)}{x+2}$
Note that virtually everything cancels: the $x-2$ terms on the left fraction, as well as a cross-canceled $x + 2$ and $x - 2$. The only part left is the $2$ in the numerator of our right-hand fraction. Thus, our answer is $2$.
Also note that this is true only if $x\ne-2, 2$, since they make the original denominators equal to 0.
