因为a^2+2a-1=0, 所以a^2+2a=1, 所以[(a-2)/(a^2+2a)-(a-1)/(a^2+4a+4)]/[(a-4)/(a+2) =[(a-2)/a(a+2)-(a-1)/(a+2)^2]*[(a+2)/(a-4)] =[(a-2)(a+2)/a(a+2)^2-a(a-1)/a(a+2)^2]*[(a+2)/(a-4)] =[(a^2-4-a^2+a)/a(a+2)^2]*[(a+2)/(a-4)] =[(a-4)/a(a+2)^2]*[(a+2)/(a-4)] =1/a(a+2) =1/(a^2+2a) =1/1 =1. 有用的话，给个好评吧O(∩_∩)O~~
试着写一下，因为不能完全确定分子分母！
{[(a^2-5a+2)/(a+2)]+1}÷[(a^2-4)/(a^2+4a+4)]
=[(a^2-5a+2+a+2)/(a+2)]×[(a^2+4a+4)/(a^2-4)]
=[(a^2-4a+4)/(a+2)]×[(a^2+4a+4)/(a^2-4)]
=[(a-2)^2/(a+2)]×[(a+2)^2/(a+2)*(a-2)]
=a-2
=(2+√3)-2
=√3.
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