Question

已知a²-3a+1=0,求a/a^4+1的值？

Answer 1

∵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),9,已知a²-3a+1=0可得：a-3+1/a=0,a+1/a=3
a/(a^4+1)
=(1/a)/(a^2+1/a^2)
=(1/a)/[(a+1/a)^2-2]
=(1/a)/7
请问下，是求a/(a^4+1)还是a^2/(a^4+1)?,1,已知a²-3a+1=0,求a/a^4+1的值

务必在今天晚上10点之前,