$A=\frac{(1-x)^2+(x+3)^2}{(x+3)(1-x)} : \frac{(x+3)^2-(x-1)^2}{(x-1)(x+3)}$
$=\frac{1-2x+x^2+x^2+6x+9}{-(x+3)(x-1)} : \frac{x^2+6x+9-x^2+2x-1}{(x-1)(x+3)}$
$=\frac{2x^2+4x+10}{-(x+3)(x-1)}.\frac{(x+3)(x-1)}{8x+8}$
$= \frac{-2(x^2+2x+5)}{2(4x+4)}$
$= \frac{-(x^2+2x+5)}{4x+4}$
Vì $-(x^2+2x+5)= -[(x+1)^2+4]<0$
$\Rightarrow 4x+4>0$
$\Leftrightarrow x>-1$