设1/2003+1/2004=x1/2003+1/2004+1/2005=y(2+1/2003+1/2004)x(1/2003+1/2004+1/2005)—(2+1/2003+1/2004+1/2005)x(1/2003+1/2004) =(2+x)y-(2+y)x=2y+xy-2x-xy=2y-2x=2(y-x)=2*1/2005=2/2005