∵a²-3a+1=0
∴a²=3a-1
则a^4=(3a-1)^2
=9a^2-6a+1
(将a²=3a-1代入)
a^4=9(3a-1)-6a+1
=21a-8
那么a^4+1=21a-8+1
=21a-7
=7(3a-1)
(将a²=3a-1代入)
则a^4+1=7a^2
所以a/(a^4+1)=a/7a^2
=1/(7a)
∵a²-3a+1=0
∴a²=3a-1
则a^4=(3a-1)^2
=9a^2-6a+1
(将a²=3a-1代入)
a^4=9(3a-1)-6a+1
=21a-8
那么a^4+1=21a-8+1
=21a-7
=7(3a-1)
(将a²=3a-1代入)
则a^4+1=7a^2
所以a/(a^4+1)=a/7a^2
=1/(7a)