(x-2)(x+2)-(x-1)(x-3)

(x-2)(x+2)-(x-1)(x-3)
=x^2-4-x^2+4x-3
=4x-7

是不等式吗？
不等式两边同乘以[(X-3)(X-2)(X-1)]^2(平方),得
(X+3)(X+2)(X+1)(X-3)(X-2)(X-1）＞0
即
(X^2-9)(X^2-4)(X^2-1)>0

则有(1)X^2-9>0,X^2-4>0,X^2-1>0,即 X<-3或X>3
(2)X^2-9<0,X^2-4<0,X^2-1>0,即 -2<X<-1或1<X<2

综上所述,不等式的解为
X<-3、-2<X<-1、1<X<2或X>3