设1/2003+1/2004+1/2005=a,1/2003+1/2004+1/2005+1/2006=b,则
原式=（2+a)b-(2+b)a
=2b+ab-2a-ab
=2(b-a)
=2(1/2003+1/2004+1/2005+1/2006-1/2003-1/2004-1/2005)
=1/1003
计算（2+1/2003+1/2004+1/2005)*(1/2003+1/2004+1/2005+1/2006)-(2+1/2003+1/2004+1/2005+1/2006)急!
计算（2+1/2003+1/2004+1/2005)*(1/2003+1/2004+1/2005+1/2006)-(2+1/2003+1/2004+1/2005+1/2006)急!
计算（2+1/2003+1/2004+1/2005)*(1/2003+1/2004+1/2005+1/2006)-(2+1/2003+1/2004+1/2005+1/2006)*(1/2003+1/2004+1/2005）
计算（2+1/2003+1/2004+1/2005)*(1/2003+1/2004+1/2005+1/2006)-(2+1/2003+1/2004+1/2005+1/2006)*(1/2003+1/2004+1/2005）
数学人气：112 ℃时间：2019-10-29 20:49:48
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