Question

代数已知a^2-4a+1=0,则(2a^5-7a^4+a^3-11a^2+7a)/(3a^2+3)的值是________

Answer 1

方法：利用a^2-4a+1=0,得到a^2-4a=-1,再把高次方降幂

2a^5-7a^4+a^3-11a^2+7a

=2a^3(a^2-4a)+a^4+a^3-11a^2+7a

=-2a^3+a^4+a^3-11a^2+7a

=a^4-a^3-11a^2+7a

=a^2(a^2-4a)+3a^3-11a^2+7a

=3a^3-12a^2+7a

=3a(a^2-4a)+7a

=4a

=a^2+1

(2a^5-7a^4+a^3-11a^2+7a)/(3a^2+3)=1/3