题目

（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)
一楼的：题没有错唉~
二楼的：

提问时间：2021-05-27

答案

1/X(X+1)=(x+1-x)/x(x+1)=(x+1)/x(x+1)-x/x(x+1)=1/x-1/(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
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/(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)