已知a^2-5a+1=0,则分式a^2/(a^4+a^2+1)的值为
问题描述：
已知a^2-5a+1=0,则分式a^2/(a^4+a^2+1)的值为
答
a^2-5a+1=0===>a^2=5a-1====>a^4=(5a-1)^2=25a^2-10a+1a^2/(a^4+a^2+1)=(5a-1)/[(25a^2-10a+1)+(5a-1)+1]=(5a-1)/(25a^2-5a+1)=(5a-1)/[25(5a-1)-5a+1)]=(5a-1)/120a-24=1/24