Question

谁能帮我解下这道数学题。谢谢了.. 下面是照片地址

Answer 1

(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。

Answer 2

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)]

追问

1/x - 1/(x+2) 这个是连除吗？

Answer 3

0.5*[1/x-1/(x+2)+1/(x+2)-1/(x+4).....+1/(x+2004)-1/(x+2006)]=1003/x(x+2006)

追问

1/x-1/(x+2) 是连除吗？

追答

不是，-那个是减号的，1/x-1/(x+2) 看不明白么？？？

Answer 4

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)

Answer 5

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) ]

Answer 6

1/2*[1/x-1/(x+2006)]