先对等式右边做通分处理,之后正好与等式左边同分母,所以只需分子对应项相等
(ax+b)/(x^2+x+1)+(cx+d)/(x^2-x+1)
=[(ax+b)(x^2-x+1)+(cx+d)(x^2+x+1)]/[(x^2+x+1)(x^2-x+1)]
=(ax^3-ax^2+ax+bx^2-bx+b+cx^3+cx^2+cx+dx^2+dx+d)/[(x^2+1)^2-x^2]
=[(a+c)x^3+(-a+b+c+d)x^2+(a-b+c+d)x+b+d]/(x^4+x^2+1)
=(6x^3+10x)/(x^4+x^2+1)
∴a+c=6……①
-a+b+c+d=0……②
a-b+c+d=10……③
b+d=0……④
联立①②③④得：a=3,b=-2,c=3,d=2