Question

两道七年级数学题~~找人帮忙~~

（1）解方程：1/X+1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3)+......+1/(X+9)(X+10)=0
（2）计算：1/(X+2)+1/(X+1)(X+3)+1/(X+2)(X+4)+......+1/(X+2007)(X+2009)
麻烦各位给个过程，详细点~~谢谢~~
另求丁博士七年级数学寒假生活20天的答案，有的唤一声~~谢(*^__^*)
一楼的：题没有错唉~~
二楼的：给个详细过程好不？

Answer 1

（1）1/X(X+1)=1/X-1/X+1,所以做皮首第一题的方程转化为1/X+(1/X-1/X+I)+(1/X+I-1/X+2)+.........+(1/X+8-1/X+9)+(1/X+9-1/X+10)=1/X+1/X-1/X+10=0.现握纤在会解了吧
（2）第二题原理一纯数样的，如1/3X5=（1/3-1/5）·1/2，最后把1/2提出去就可以了

Answer 2

<1>1/X(X+1)=(x+1-x)/巧颂隐x(x+1)=(x+1)/x(x+1)-x/x(x+1)=1/x-1/孝厅(x+1)
1/(X+1)(X+2)=[x+2-(x+1)]/(x+1)(x+2)=1/(x+1)-1/(x+2)
.............
依次类推
1/X+1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3)+......+1/(X+9)(X+10)
=1/x+1/x-1/(x+1)+1/(x+1)-1/(x+2)+.......+1/(x+9)-1/(x+10)
=2/x-1/(x+10)=0
x=2x+20,
x=-20

<2>方法同1，1/(X+1)(X+3)=0.5*[（x+3）-(x+1)/(x+1)(x+3)]=0.5[1/(x+1)-1/(x+3)]
1/(X+2)(X+4)=0.5*[(x+4)-(x+2)/(x+2)(x+4)]=0.5[1/(x+2)-1/(x+4)]
...........
1/樱歼(X+2)+1/(X+1)(X+3)+1/(X+2)(X+4)+......+1/(X+2007)(X+2009)
=1/(x+2)+0.5[1/(x+1)-1/(x+3)+1/(x+2)-1/(x+4)+1/(x+3)-1/(x+5)+.....+1/(x+2007)-1/(x+2009)]
=1/(x+2)+0.5[1/(x+1)+1/(x+2)-1/(x+2008)-1/(x+2009)]
=1/(2x+2)+3/(2x+4)+-1/(2x+4016)-1/(2x+4018)

Answer 3

1、1/X+1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3)+......+1/(X+9)(X+10)=0
变形为：1/x+1/x - 1/(x+1)+1/丛尘(x+1)-1/(x+2)+……+1/(X+9) - 1/(X+10)=0
整理首祥，得2/x- 1/(X+10)=0
得x=-20
2、1/(X+2)+1/(X+1)(X+3)+1/(X+2)(X+4)+......+1/(X+2007)(X+2009)
=1/(X+2)+1/2[1/(X+1) - 1/(X+3)+1/(X+2) - 1/(X+4)+......+1/(X+2007) - 1/(X+2009)]
=1/(X+2)+1/2[1/(X+1)+1/(X+2)- 1/(x+2008)- 1/(X+2009)]
=3/2*1/(x+2)+1/2[1/(X+1)- 1/(x+2008)- 1/(X+2009)]
这个答渗芹禅案绝对没问题

Answer 4

(1)左边=1-1/源模x+1/x-1/(x+1)+1/(x+1)...-1/(x+9)+1/(x+9)-1/(x+10)
=1-1/(x+10)
变化后得方程：1=1/(x+10)
同时旦侍乘(x+10)得 x+10=1
x=9
(2)
1/X(X+2)+1/雹迟缓(X+1)(X+3)+1\(X+2)(X+4)+......+1/(X+2007)(X+2009)
=(1/2)(1/x-1/(x+2)+1/(x+2)-1/(x+4)+1/(x+4)....1/(x+2007)-1/(x+2009))
=(1/2)*(1/2-1/(x+2009))
=(x+2007)/4(x+2009)

Answer 5

给你个提示就很容易了，1/X(X+1)=1/x-1/(x+1)，后面灶数亏的几个都毕歼保持队形展开隐神，能前后抵消
两题能简便运算 ，第2题还要在后面几项前面乘1/2,能抵消就对了