高手赐教，数学题，马上要过程和答案

a^2+2a-1=0
所以a^2+2a=1

原式=[(a-2)/a(a+2)-(a-1)/(a+2)^2]*[(a+2)/(a-4)]
=[(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)]
=[(a-2)(a+2)-a(a-1)]/[a(a+2)(a-4)]
=(a^2-4-a^2+a)/[a(a+2)(a-4)]
=(a-4)/[(a^2+2a)(a-4)]
=1/(a^2+2a)
=1/1
=1

[（a-2/a^2+2a)-(a-1/a^2+4a+4)]/(a-4/a+2)
=(a-2/a^2+2a-a+1/a^2-4a-4)/(a-4/a+2)
=(-2a-1/a^2-4)/(a-4/a+2)
=-(2a^3+4a^2+1)/(a^2+2a-4)
=-[2a(a^2+2a)+1]/(1-4)
=(2a+1)/3

[（a-2/a^2+2a)-(a-1/a^2+4a+4)]/(a-4/a+2)
=[(a-2)(a+2)-a(a-1)]*(a+2)/a(a+2)^2(a-4)
=(a-4)(a+2)/a(a+2)^2(a-4)
=(a-4)/a(a+2)(a-4)=1/a(a+2)=1/(a^2+2a)=1
很高兴为你解决问题！