来自李金屏的问题
计算a−ba÷(a−2ab−b2a)的结果是()A.1a+bB.−1a+bC.1a−bD.−1a−b
计算a−ba÷(a−2ab−b2a)的结果是()
A.1a+b
B.−1a+b
C.1a−b
D.−1a−b


计算a−ba÷(a−2ab−b2a)的结果是()A.1a+bB.−1a+bC.1a−bD.−1a−b
计算a−ba÷(a−2ab−b2a)的结果是()
A.1a+b
B.−1a+b
C.1a−b
D.−1a−b
原式=a−ba