a^2/(a^4+1)=1/[(a^4+1)/a^2]
(a^4+1)/a^2=a^2+1/a^2
所以：a^2/(a^4+1)=1/[a^2+1/a^2].1
a^2-3a+1=0
除以a,
a-3+1/a=0
a+1/a=3
平方：
a^2+1/a^2+2=3
a^2+1/a^2=1代入1式：
a^2/(a^4+1)=1/[a^2+1/a^2]=1/1=1正確答案是1/7，我不會(huì)做，但我只知道得數(shù)，你確定沒做錯(cuò)？是做錯(cuò)了。a^2/(a^4+1)=1/[(a^4+1)/a^2](a^4+1)/a^2=a^2+1/a^2所以：a^2/(a^4+1)=1/[a^2+1/a^2]...............1a^2-3a+1=0除以a,a-3+1/a=0a+1/a=3平方：a^2+1/a^2+2=9a^2+1/a^2=7代入1式：a^2/(a^4+1)=1/[a^2+1/a^2]=1/7=1 /7有沒有簡便的方法呀。不太懂！最簡單方法：a^2-3a+1=0a^2=3a-1代入：a^2/(a^4+1)=(3a-1)/[(3a-1)^2+1]=(3a-1)/(9a^2-6a+2)=(3a-1)/(9(3a-1)-6a+2)=(3a-1)/(27a-9-6a+2)=(3a-1)/(21a-7)=(3a-1)/[7(3a-1)]=1/7