来自莫炳禄的问题
【分式子通分*((x^2-2x+1)/(x^2-1))/((x-1)/(x^2+x))*(4/(a+2))+a-2+a-2[(a+2)(a-2)/(a+2)]】
分式子通分
*((x^2-2x+1)/(x^2-1))/((x-1)/(x^2+x))
*(4/(a+2))+a-2
+a-2
[(a+2)(a-2)/(a+2)]


【分式子通分*((x^2-2x+1)/(x^2-1))/((x-1)/(x^2+x))*(4/(a+2))+a-2+a-2[(a+2)(a-2)/(a+2)]】
分式子通分
*((x^2-2x+1)/(x^2-1))/((x-1)/(x^2+x))
*(4/(a+2))+a-2
+a-2
[(a+2)(a-2)/(a+2)]
(1)[(x^2-2x+1)/(x^2-1)]/[(x-1)/(x^2+x)]
=(x-1)^2/[(x+1)(x-1)]*{[x(x+1)]/(x-1)}
=(x-1)/(x-1)*x
=x
(2)[4/(a+2)]+a-2
=[4/(a+2)]+[(a+2)(a-2)/(a+2)]
=a^2/(a+2)