一道数学题 必有过程

来源：百度知道 编辑：UC知道 时间：2024/06/04 08:10:17
a^2-2a+1/a^2-1除以a-1/a^2+a

2x/x-3加6/3-x

x+2减4/2-x

-1+2a-a^2/a+2乘a^2+2a/a^2-1

[(a^2-2a+1)/(a^2-1)]÷[(a-1)/(a^2+a)]
=[(a-1)^2/(a+1)(a-1)]÷[(a-1)/a(a+1)]
=[(a-1)^2/(a+1)(a-1)]*[a(a+1)/(a-1)]
=(a-1)^2a(a+1)/(a+1)(a-1)^2
=a

2x/(x-3)+6/(3-x)
=2x/(x-3)-6/(x-3)
=(2x-6)/(x-3)
=2(x-3)/(x-3)
=2

x+2-4/(2-x)
=[(x+2)(2-x)-4]/(2-x)
=(2-x^4-4)/(2-x)
=-x^2/(2-x)
=x^2/(x-2)

[(-1+2a-a^2)/(a+2)]*[(a^2+2a)/(a^2-1)]
=[-(a-1)^2/(a+2)]*[a(a+2)/(a+1)(a-1)]
=-(a-1)^2a(a+2)/[(a+2)(a+1)(a-1)]
=-a(a-1)/(a+1)
=(a-a^2)/(a+1)