X(X+z)分之1+(X+1)(X+2)分之1+.+(x+8)(x+9)分之1=x9(x+9)分之2x+3
问题描述：
X(X+z)分之1+(X+1)(X+2)分之1+.+(x+8)(x+9)分之1=x9(x+9)分之2x+3
答
1/x(x+1)1/(x+1)(x+2)+.+1/(x+8)(x+9)=(2x+3)/x(x+9)1/x-1/(x+1)+1/(x+1)-1/(x+2)+.+1/(x+8)-1/(x+9)=(2x+3)/x(x+9)1/x-1/(x+9)=(2x+3)/x(x+9)(x+9-x)/x(x+9)=(2x+3)/x(x+9)9/x(x+9)=(2x+3)/x(x+9)9/x(x+9)=(2x+3)/x...有简单点的方法么已经很简单了看不懂 ，1/x(x+1)1/(x+1)(x+2)+......+1/(x+8)(x+9)=(2x+3)/x(x+9)1/x-1/(x+1)+1/(x+1)-1/(x+2)+.....+1/(x+8)-1/(x+9)=(2x+3)/x(x+9)等么1/x(x+1)+1/(x+1)(x+2)+......+1/(x+8)(x+9)=(2x+3)/x(x+9)1/x(x+1)=1/x-1/(x+1)1/(x+1)(x+2)=1/(x+1)-1/(x+2)..........1/(x+8)(x+9)=1/(x+8)-1/(x+9)