因为(x的平方-2x+1)/(x的平方-1)÷(x-1)/(x的平方+x)-x的值与x无关:
(x的平方-2x+1)/(x的平方-1)÷(x-1)/(x的平方+x)-x
=(x-1)^2/[(x+1)(x-1)] * x(x+1)/(x-1) -x
=x -x
=0
因为(x的平方-2x+1)/(x的平方-1)÷(x-1)/(x的平方+x)-x的值与x无关:
(x的平方-2x+1)/(x的平方-1)÷(x-1)/(x的平方+x)-x
=(x-1)^2/[(x+1)(x-1)] * x(x+1)/(x-1) -x
=x -x
=0