S=

1

(1+x)2

+

1

(1-x)2

=

(1-x)2+(1+x)2

(1+x)2(1-x)2

=

1-2x+x2+1+2x+x2

(1-x2)2

=

2+2x2

(1-x2)2

则S-2=

2+2x2

(1-x2)2

-2=

2+2x2-2+4x2-2x4

(1-x2)2

=

6x2-2x4

(1-x2)2

=

2x2(3-x2)

(1-x2)2

=

2x2

(1-x2)2

×(3-x2)

由于

2x2

(1-x2)2

≥0，（3-x²）不能确定它的符号

S-2可能大于零也可能小于零，还可能等于零．

故选D