实数a满足a2+2a-1=0,求(a-2/a2+2a-a-1/a2+4a+4)/a-4/a+2的值

∵a²+2a-1=0
∴a²+2a=1
(a-2)/(a²+2a)-(a-1)/(a²+4a+4)-(a-4)/(a+2)....[猜着修改的]
=(a-2)/(a²+2a)-(a-1)/(a²+4a+4)-(a-4)/(a+2)
=(a-2)/a(a+2)-(a-1)/(a+2)²-(a-4)/(a+2)
=(a²-4)/a(a+2)²-a(a-1)/a(a+2)²-(a-4)/(a+2)
=(a-4)/a(a+2)²-(a-4)/(a+2)
=(a-4)[1/a(a+2)²-(a²+2a)/a(a+2)²]
=(a-4)[1/a(a+2)²-1/a(a+2)²]
=(a-4)×0
=0

10.先化简，再求值：
((a-2/a²+2a)-(a-1/a²+4a+4))÷(a-4/a+2)，其中a满足：a²+2a-1=0.

((a-2/a(a+2))-(a-1/(a+2)²)×(a+2/a-4)
=((a²-4/a(a+2)²)-(a(a-1)/a(a+2)²))×(a+2/a-4)
=(a²-4-a²+a/a)×(1/a-4)
=1/a
解方程：a²+2a-1=0
a=1 b=2 c=-1
b²-4ac=2²-4×1×(-1)=8
a¹=(-2+2√2)/2=√2-1
a²=(-2-2√2)/2=-√2-1
当a=a¹时1/a=1/√2-1
当a=a²时1/a=-(1/√2+1)
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