x的平方+4x+4分之1-x的平方除以（x-1）的平方乘以(x-1)分之x的平方+3x+2

(1-x)^2/(x^2+4x+4)/(x-1)^2*(x^2+3x+2)/(x-1)
=(x+1)/(x+2)(x-1)
=(x+1)/(x^2+x-2)是1-x的平方，不是（1-x）的平方原式=(1-x)(1+x)/(x+2)(x+2)/(1-x)^2*(x+1)(x+2)/(x-1)=-(x+1)(x+1)/[(x+2)(x-1)(x-1)]=-(x^2+2x+1)/(x^3-3x+2)