已知a²-3a+1=0,则4a²-9a-2+9/(1+a²)=?

问题描述:

已知a²-3a+1=0,则4a²-9a-2+9/(1+a²)=?

即a²+1=3a
a²=3a-1
所以原式=4(3a-1)-9a-2+9/3a
=3a-6+3/a
=(3a²+3)/a-6
=3(a²+1)/a-6
=3*3a/a-6
=9-6
=3