先化简,

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先化简,
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先化简,
先化简,

先化简,
a^2+2a-1=0
a^2+2a=1
原式=[(a-2)/a(a+2)-(a-1)/(a+2)^2]*(a+2)/(a-4)
==[(a-2)(a+2)/a(a+2)^2-(a-1)a/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

  题能大一点吗?