(1+x^2)/(1+x^6)=1/(x^4-x^2+1)=1/[x^2+3^(1/2)x+1]*[x^2-3^(1/2)x+1]设原式=(ax+b)/[x^2+3^(1/2)x+1]+(cx+d)/[x^2-3^(1/2)x+1]a+c=0 b+d+3^(1/2)(c-a)=0 a+c+3^(1/2)(d-b)=0b+d=1则a=3^(1/2)/6 c=-3^(1/2)/6 b=d=1/...