因为 x/(x^2+x+1) =1/5,所以分子x不等于0,（因为当x=0时,代入0/1=0,不等于1/5）
即 5x=x^2+x+1
x^2=4x-1,因为x不等于0,所以x^2>0,即4x-1>0
所求式子为 (x^2)/(x^4+x^2+1)=x/[(x^2)^2+x^2+1]
把x^2=4x-1代入得 (4x-1)/[(4x-1)^2+4x-1+1]
=(4x-1)/[(4x-1)^2+4x-1+1]
=(4x-1)/[(4x-1)^2+4x]
=(4x-1)/(16x^2-8x+1+4x)
=(4x-1)/(16x^2-4x+1)
再把x^2=4x-1代入得 (4x-1)/[16（4x-1)-4x+1]
= (4x-1)/[16（4x-1)-(4x-1)]
= (4x-1)/[15（4x-1)]
又因为 4x-1>0 ,所以原式等于1/15