先化简，再求值【初三数学】

显示全部楼层 · 2013-1-31 16:28:21

a^2+2a-1=0a^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...

千问 · 2013-1-31 16:28:21

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千问 · 2013-1-31 16:28:21

题能大一点吗？...