【数学分式计算,要过程】a²分之（a²-ab）÷（b分之a-a分之ba²分之（a²-ab）÷（b分之a-a分之b）（a²-1）分之（a²+2a+1）-（a-1）分之1
问题描述：
【数学分式计算,要过程】a²分之（a²-ab）÷（b分之a-a分之b
a²分之（a²-ab）÷（b分之a-a分之b）
（a²-1）分之（a²+2a+1）-（a-1）分之1
答
a²分之（a²-ab）÷（b分之a-a分之b）
=(a-b)/a÷(a-b)(a+b)/ab
=[(a-b)/a]xab/[(a-b)(a+b)]
=b/(a+b)
（a²-1）分之（a²+2a+1）-（a-1）分之1 (
=[(a+1)(a+1)/(a-1)(a+1)]-[1/(a-1)]
=(a+1)/(a-1)-[1/(a-1)]
=a+1-1
=a
答
a²分之（a²-ab）÷（b分之a-a分之b）
=a(a-b)/a^2/[(a^2-b^2)/ab]
=(a-b)/a*ab/(a+b)(a-b)
=b/(a+b)
（a²-1）分之（a²+2a+1）-（a-1）分之1
=(a+1)^2/(a+1)(a-1)-1/(a-1)
=(a+1)/(a-1)-1/(a-1)
=a/(a-1)
答
1、=a(a-b)/a²÷(a²-b²)/ab
=(a-b)/b×ab/(a+b)(a-b)
=a/(a+b)
2、=(a+1)²/(a+1)(a-1)-1/(a-1)
=(a+1)/(a-1)-1/(a-1)
=1/(a-1)