1/x(x+1)=[(x+1)-x]/x(x+1)=(x+1)/x(x+1)-x/x(x+1)=1/x-1/(x+1)同理1/(x+1)(x+2)=[(x+2)-(x+1)]/(x+1)(x+2)=1/(x+1)-1/(x+2)以此类推原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+……+1/(x+8)-1/(x+9)中间正负抵消=1/x-1/(x+9...