题目

已知a^2-3a+1=0,求a+1/a,a^2+1/a^2,a^2-1/a^2,a^3+1/a^3,4a^2-9a-2+9/(1+a^2)

提问时间：2020-11-06

答案

∵a^2-3a+1=0 ∴a-3+1/a=0 ∴a+1/a=3
∴(a+1/a)^2=a^2+2+1/a^2=9 ∴a^2+1/a^2=7
∴a^2+1/a^2-2=(a-1/a)^2=5 ∴a-1/a=±√5
∴a^3+1/a^3=(a+1/a)(a^2+1/a^2-1)=3×(7-1)=18
∵a^2-3a+1=0 ∴a^2+1=3a,a^2=3a-1
∴(4a^2-9a-2)+9/(1+a^2)=4(3a-1)-9a-2+9/3a
=3a-6+3/a
=3(a+1/a)-6
=3
∴a+1/a=3,a^2+1/a^2=7,a-1/a=±√5,a^3+1/a^3==18,(4a^2-9a-2)+9/(1+a^2)=3