一道数学问题~要列式啊!!
简便算法<(1+1/1999)+(1/2000)+(1/2001)>*<(1/1999)+(1/2000)+(1/2001)+(1/2002)>-<(1+1/1999)+(1/2000)+(1/2001)+(1/2002)>*<(1/1999)+(1/2000)+(1/2001)>
参考答案:设A=1/1999+1/2000+1/2001
原式=[A+1]*[A+1/2002]-[1+A+1/2002]*A
=A^2+A/2002+A+1/2002-[A+A^2+A/2002]
=1/2002