已知a^2-5a+1=0,则分式a^2/(a^4+a^2+1)的值为

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/16 04:24:42

已知a^2-5a+1=0,则分式a^2/(a^4+a^2+1)的值为
已知a^2-5a+1=0,则分式a^2/(a^4+a^2+1)的值为

已知a^2-5a+1=0,则分式a^2/(a^4+a^2+1)的值为
a^2-5a+1=0===>a^2=5a-1====>a^4=(5a-1)^2=25a^2-10a+1
a^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