(1/2+1/3+...+1/2005)(1+1/2+1/3+...+1/2004)-(1+1/2+...+1/2005)(1/2+1/3+...+1/2004)不用设未知数怎么解?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/09 07:40:57
![(1/2+1/3+...+1/2005)(1+1/2+1/3+...+1/2004)-(1+1/2+...+1/2005)(1/2+1/3+...+1/2004)不用设未知数怎么解?](/uploads/image/z/6897693-21-3.jpg?t=%281%2F2%2B1%2F3%2B...%2B1%2F2005%29%281%2B1%2F2%2B1%2F3%2B...%2B1%2F2004%29-%281%2B1%2F2%2B...%2B1%2F2005%29%281%2F2%2B1%2F3%2B...%2B1%2F2004%29%E4%B8%8D%E7%94%A8%E8%AE%BE%E6%9C%AA%E7%9F%A5%E6%95%B0%E6%80%8E%E4%B9%88%E8%A7%A3%3F)
xR=O@+Ҵմ8 hWZZ)E'
1㨋 ?Ja0垯{BX'm6Ҧeϛ{6{|Ln'_@&iNEQLN1Pa-] y
"eF"Ix>i"WX#.tK TJP>KAgṉP\̃>㼬7B.ުsI|
T1ci\nإ\[6~F6ٖ6LtXL\5ČGC̳mh<293\fOURE4?
(1/2+1/3+...+1/2005)(1+1/2+1/3+...+1/2004)-(1+1/2+...+1/2005)(1/2+1/3+...+1/2004)不用设未知数怎么解?
(1/2+1/3+...+1/2005)(1+1/2+1/3+...+1/2004)-(1+1/2+...+1/2005)(1/2+1/3+...+1/2004)不用设未知数怎么解?
(1/2+1/3+...+1/2005)(1+1/2+1/3+...+1/2004)-(1+1/2+...+1/2005)(1/2+1/3+...+1/2004)不用设未知数怎么解?
=[(1/2+1/3+...+1/2004)+1/2005][1+(1/2+1/3+...+1/2004)]-[1+(1/2+1/3+...+1/2004)+1/2005](1/2+1/3+...+1/2004)
=(1/2+1/3+...+1/2004)×1+(1/2+1/3+...+1/2004)^2+1/2005+](1/2+1/3+...+1/2004)×1/2005-1×(1/2+1/3+...+1/2004)-(1/2+1/3+...+1/2004)^2=1/2005×(1/2+1/3+...+1/2004)
=1/2005