各分式减1,得1/（x+2004)+1/(x+2006)=1/(x+2007)+1/(x+2003),通分得（2x+4010)/[（x+2004)(x+2006)]=(2x+4010)/[(x+2003)(x+2007)],两边的分母不等,∴2x+4010=0,x=-2005.检验知,它是原方程的根....