.下面是照片地址
来源：学生作业帮助网 编辑：六六作业网 时间：2025/01/12 05:32:23
.下面是照片地址
.下面是照片地址
.下面是照片地址
(1/2) [ 1/x - 1/(x+2) + 1/(x+2) - 1/(x+4) + 1/(x+4) - 1/(x+6) + ...+1/(x+2004) - 1/(x+2006)]
=(1/2) [ 1/x - 1/(x+2006)]
=1003/[x(x+2006)]
解析：根据原题给出的解法,你可以做出大胆的推理,即1/x*1/(x+2)能否拆成题干所给的形式.所以你就可以想1/x和1/(x+2)能不能拆成“?”的形式.然后你就会发现可以把这两个式子相减再除以2和原式相等.接着依此类推.再结合题干的计算方法,发现除了第一项和最后一项外,其余的都能合并,化0.所以只需把1/2提出来,首相和末相相减.即可得出答案.
实在是不懂，要不就帮你了
0.5*[1/x-1/(x+2)+1/(x+2)-1/(x+4).....+1/(x+2004)-1/(x+2006)]=1003/x(x+2006)
1/2*[1/x-1/(x+2006)]
原式=(1/2) [ 1/x - 1/(x+2) + 1/(x+2) - 1/(x+4) + 1/(x+4) - 1/(x+6) + ... +1/(x+2004) - 1/(x+2006)]
=(1/2) [ 1/x - 1/(x+2006)]
=1003/[x(x+2006)]
原式=(1/2)[ 1/x-1/(x+2)+1/(x+2)-1/(x+4)+1/(x+4)-1/(x+6)+·······+1/(x+2004)-1/(x+2006) ]
=(1/2)[1/x -1/(x+2006)]
=1003/ [ x(x+2006) ]
类比运用嘛！
=1/2 x {1/x -1/(x+2) +1/(x+2) - 1/(x+4)+.....+1/(x+2004) -1/(x+2006) }
=1/2 x { 1/x - 1/(x+2006) }
=1003/x(X+2006)