请教一道分式的题！急求~~

来源：百度知道 编辑：UC知道 时间：2024/05/29 22:32:04
这道题是：

1/x（x+1）+ 1/(x+1)(x+2) + 1/(x+2)(x+3)+...+1/(x+8)(x+9)
化简求值，

快！

谢谢

1/x（x+1）+ 1/(x+1)(x+2) + 1/(x+2)(x+3)+...+1/(x+8)(x+9)
=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]+...+[1/(x+8)-1/(x+9)]
=1/x-1/(x+9)
=9/[x(x+9)]

1/[x(x+1)]=1/x - 1/(x+1)
以此类推
原式=1/x - 1/(x+9)

1/x（x+1）+ 1/(x+1)(x+2) + 1/(x+2)(x+3)+...+1/(x+8)(x+9)
=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]+...+[1/(x+8)-1/(x+9)]
=1/x-1(x+1)+1/(x-1)-1/(x+2)~~~~+1/(x+8)-1/(x+9) 去完括号如此，在前后约分得：
=1/x-1/(x+9)
=9/[x(x+9)]

裂项相消法 =(x+1)/x+1/(x+1)-1/(x+2)+.....+1/(x+8)-1/(X+9)=(1+x)/x-1/(x+9)=(x+3)的平方/x与(x+9)的积