已知a2-5a+1=0,求a4+1a2的值.

问题描述:

已知a2-5a+1=0,求

a4+1
a2
的值.

由已知a2-5a+1=0得a≠0,则将已知等式两边同除以a得a-5+

1
a
=0,
∴a+
1
a
=5,
a4+1
a2
=a2+
1
a2

=(a+
1
a
2-2
=52-2
=23.