( 1-a)^2- 1=a^2-2a<； =0, that is, 0

At a<0 or a>2, let x1= a-1-√ (a 2-2a), and x2 = a-1+√ (a 2-2a).

X 12。 By x>0 and x≠ 1, lnx/(x-1) >; A/(x+ 1) holds, so.

A< (x+1) lnx/(x-1), denoted as h(x), then

h'(x)={(x- 1)[lnx+(x+ 1)/x]-(x+ 1)lnx}/(x- 1)^2

=(x- 1/x-2lnx)/(x- 1)^2，

Let f (x) = x- 1/x-2lnx, x >；; 0, then f' (x) = (x- 1) 2/x > =0,∴F(x) is increasing function and F( 1)=0.

0x>F (x) at 1 >: 0, h'(x)>0, h(x) is increasing function,

When x→ 1, h(x)→lnx+(x+ 1)/x (Robida's Law) →2,

∴a<； =2, do whatever you want.