分式方程?

来源：百度知道 编辑：UC知道 时间：2024/06/06 07:28:43
[1/(X-1)X]+[1/X(X+1)]+[1/(X+1)(X+2)]+ ……… +[1/(X+9)(X+10)]=[11/12(X-1)]

左边=1/(X-1)-1/X+1/X-1/(X+1)+1/(X+1)-1/(X+2)+……+1/(X+9)-1/(X+10)
=1/(X-1)-1/(X+10)=11/[12(X-1)]
所以1/[12(X-1)]=1/(X+10)
12(X-1)=X+10
12X-12=X+10
X=2

原方程即为[(x-(x-1))/(x-1)x]+[((x+1)-x)/x(x+1)]+……+[((x+10)-(x+9))/(x+9)(x+10)]=[11/12(X-1)]
即1/(x-1)-1/x+1/x-1/(x+1)+1/(x+1)-1/(x+2)+……+1/(x+9)-1/(x+10)=[11/12(X-1)]
得1/(x-1)-1/(x+10)=[11/12(X-1)]
x+10-(x-1)=11(x+10)/12
解得x=2