Question

先化简,再求值:[(a的平方+2a分之a-2)-(a的平方+4a+4分之a-1)除以(a+2分之a-4),其中a=1

把后面那个除以的式子分别乘进去的！分配律！急死了啊！！！！

Answer 1

[(a-2)/(a²+2a)-(a-1)/(a²+4a+4)]÷(a-4)/(a+2)
=(a-2)/(a(a+2))*(a+2)/(a-4)-(a-1)/(a+2)²*(a+2)/(a-4)
=(a-2)/(a(a-4))-(a-1)/((a+2)(a-4))
=(1-2)/(1*(1-4))-(1-1)/((1+2)(1-4))
=-1/(-3)-0
=1/3

追问

第三步没看明白。。。求解释一下

追答

(a-2)/(a(a+2))×(a+2)/(a-4)
=(a-2)(a+2)/(a(a+2)(a-4))

=(a-2)/(a(a-4))
(a-1)/(a+2)²*(a+2)/(a-4)

=(a-1)(a+2)/((a+2)²(a-4))
=(a-1)/((a+2)(a-4))