$x>2=>x-2>0\\ P=\left(\dfrac{x}{x+1}-\dfrac{1}{1-x}+\dfrac{1}{1-x^2}\right):\dfrac{x-2}{x^2-1}\\ =\dfrac{x^2}{x-2}\\ =\dfrac{(x^2-4x+4)+(4x-8)+4}{x-2}\\ =\dfrac{(x-2)^2}{x-2}+\dfrac{(4x-8)}{x-2}+\dfrac{4}{x-2}\\ =x-2+4+\dfrac{4}{x-2}\\ =(x-2)+\dfrac{4}{x-2}+4\\ \ge 2\sqrt{(x-2).\dfrac{4}{x-2}}+4(BĐT \, Cauchy)=8(đpcm)$
Dấu "=" xảy ra
$<=> x-2=\dfrac{4}{x-2}<=>x=4$