Question

求解，初二数学分式

Answer 1

解 原式=【1/x-1/(x+1)】+【1/(x+1)-1/(x+2)】+【1/(x+2)-1/(x+3)】....+【1/(X+2008)-1/(X-2009)】
=1/x-1/(X+2009)
或者)=（x+2008）/【x（x+2009）】
你可以用实数对比1/【3（3+1）】+1/【（3+1）（3+2）】=1/3*4+1/4*5=1/3-1/4+1/4-1/5 这样就清楚

Answer 2

1/[x(x+1)]=1/x-1/(x+1)
1/[(x+1)(x+2)]=1/(x+1)-1/(x+2)
.......
1[/(x+2008)(x+2009)]=1/(x+2008)-1/(x-2009)
所以原式＝1/x-1/(x+2009)=2009/[x(x+2009)]

Answer 3

由1/（x+n）(x+n+1)= 1/（x+n）-1/(x+n+1)
所以 原式= 1/x -1/（x+1）+1/（x+1)-1/(x+2)+1/(x+2)-1/(x+3)……+1/(x+2008)-1/(x+2009)
=1/x-1/(x+2009)=（x+2008）/x（x+2009）

Answer 4

1/x(x+1)=1/x-1/(x+1)
依次类推
希望能帮到你

Answer 5

1/x(x+1)=1/x-1/(x+1)
依次类推
希望能帮到你

Answer 6