解：
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.
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