有一个初二的解方程怎么算:[(a+1)\(a-1)+1\(a-1)²]×a\a-1

有一个初二的解方程怎么算:[(a+1)\(a-1)+1\(a-1)²]×a\a-1
有一个初二的解方程怎么算:[(a+1)\(a-1)+1\(a-1)²]×a\a-1

有一个初二的解方程怎么算:[(a+1)\(a-1)+1\(a-1)²]×a\a-1
[(a+1)\(a-1)+1\(a-1)²]×a\a-1
=【(a+1)(a-1)+1】\(a-1)²xa\a-1
=a²\(a-1)²xa\a-1
=a3/(a-1)3
a的3次方/（a-1）的3次方
记得给满意啊

=[(a+1)(a-1)-(a-1)]/(a-1)(a-1)*a/a-1 =a3/(a-1)3 题目有点怪,没错吧?

a3/(a-1)3