x^2-3x+1=0
x+(1/x)=3
x^2/(x^4+x^2+1)
=1/[x^2+(1/x^2)+1]
=1/[(x+1/x)^2-1]
原式＝1/[(x+1/x)^2-1]
＝1/(3^2-1)
=1/8
x^4+x^2+1
=(3x-1)^2+x^2+1
=10x^2-6x+2
=10(x^2-3x+1) +24x-8
=8(3x-1)
x^2/(x^4+x^2+1)
=(3x-1)/[8(3x-1)]
=1/8
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