作业帮已知1/1×2=1-1/2,1/2×3=1/2-1/3,1/3×4=1/3-1/4.,则1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/18 19:31:27
x){}K
O7557Ӎmk n&z:O;fWhThjji a8ƚ6IE43[Ά.'Lhv.T
L.TL95<;wl}'X @1.hoѪf A>̊l풗f<ݰٌO;V>Yw˅;_có';־(|'&=l[Df<H5O+VHɵC! jO
已知1/1×2=1-1/2,1/2×3=1/2-1/3,1/3×4=1/3-1/4.,则1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)
已知1/1×2=1-1/2,1/2×3=1/2-1/3,1/3×4=1/3-1/4.,则1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)
已知1/1×2=1-1/2,1/2×3=1/2-1/3,1/3×4=1/3-1/4.,则1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)
=1/x-1/(x+3)
原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)
=1/x-1/(x+3)
=3/<(x+3)*x>
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)=1/x-1/(x+3)=3/[x(x+3)]
此题就是利用拆项达到把中间项相互抵消消项的目的,1/n*m=(1/n-1/m)/(m-n),m>n