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_________.注：
问题描述：
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_________.
注：
答
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
答
2a^5-7a^4+a^3-11a^2+7a
=(3a+a^2+2a^3)(a^2-4a+1)+4a
=4a
=a^2-4a+1+4a
=a^2+1
(2a^5-7a^4+a^3-11a^2+7a)/(3a^2+3)=(a^2+1)/(3a^2+3)=3
答
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