已知y=(x2+2x+1)\(x2-1)/(x+1)\(x2-x)-x+1.试说明不论x为何值,y的值不变

已知y=(x2+2x+1)\(x2-1)/(x+1)\(x2-x)-x+1.试说明不论x为何值,y的值不变
已知y=(x2+2x+1)\(x2-1)/(x+1)\(x2-x)-x+1.试说明不论x为何值,y的值不变

已知y=(x2+2x+1)\(x2-1)/(x+1)\(x2-x)-x+1.试说明不论x为何值,y的值不变
y=[(x^2+2x+1)/(x^2-1)]/[(x+1)/(x^2-x)]-x+1
=(x+1)^2/(x-1)(x+1)/[(x+1)/x(x-1)]-x+1
=(x+1)/(x-1)*x(x-1)/(x+1)-x+1
=x-x+1
=1
所以不论x为何值,y的值不变 ,不过x≠1且x≠-1且x≠0

y=(x+1)\(x-1)/(x+1)\(x-1)*x-(x-1)
=x-x+1
=1