先化简,再求值:[(a^2-4)\(a^2-4a+4)-a/(2-a)]÷2/(a^2-2a),其中,a是方程x^2+3x+1=0的根

[(a^2-4)/(a^2-4a+4)-a/(2-a)]÷[2/(a^2-2a)]
=[(a+2)(a-2)/(a-2)^2-a/(2-a)]*[(a^2-2a)/2]
=[(a+2)/(a-2)-a/(2-a)]*[(a^2-2a)/2]
=[2/(a-2)]*[a(a-2)/2]
=a
a是方程x^2+3x+1=0的解,
a^2+3a+1=0,
a=(-3±√5)/2
a^2+3a+1=0
a^2+2(3/2)a+(3/2)^2+1-9/4=0
(a+3/2)^2=5/4
a+3/2=±√5/2
a=-3/2±√5/2