已知a^2-4a+1=0,求[2a^5-7a^4+a^3-11a^2+7a]/[3a^2+3]
问题描述：
已知a^2-4a+1=0,求[2a^5-7a^4+a^3-11a^2+7a]/[3a^2+3]
答
a^2-4a+1=0,a^2+1=4a
2a^5-7a^4+a^3-11a^2+7a
=2a^3(a^2-4a+1)+a^2(a^2-4a+1)+3a(a^2-4a+1)+4a
=0+0+0+4a
=4a
3a^2+3=3(a^2+1)=3*4a
原式=4a/12a=1/3 ,a≠0