b) $B=\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}-\frac{1}{x-2}\right).(x^2-4)$.

b) $B=\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right).(x^2-4)$

$=\left[\frac{1}{{(x+2)}^2}-\frac{1}{{(x-2)}^2}\right]:\left[\frac{x-2}{(x+2)(x-2)}+\frac{x+2}{(x+2)(x-2)}\right].\left[\right(x+2\left)\right(x-2\left)\right]$

$=\frac{{(x-2)}^2-{(x+2)}^2}{{(x+2)}^2{(x-2)}^2}:\frac{(x-2)+(x+2)}{(x+2)(x-2)}.\left[\right(x+2\left)\right(x-2\left)\right]$

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

$=\frac{-8x}{x-2+x+2}=\frac{-8x}{2x}=-4$.

Do đó, giá trị của biểu thức B không phụ thuộc vào giá trị của biến.