Question

解分式方程1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)=1/8

Answer 1

将1/(x+1)(x+2)拆开以此类推1/(x +2)-1/(x +1)就行了，剩下的也是这样

Answer 2

1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)=1/8
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)=1/8
1/(x+1)-1/(x+5)=1/8
32=(x+1)(x+5)
x²+6x-27=0
x=-9 x=3
检验

Answer 3

裂项
解：1/（x+1）-1/（x+2）+1/（x+2）-1/（x+3）+1/（x+3）-1/（x+4）+1/（x+4）-1/（x+5）=1/8
1/（x+1）-1/（x+5）=1/8
4/（x+1)（x+2）=1/8
x=3

Answer 4

拆项相消：
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)=1/8
即：1/(x+1)-1/(x+5)=1/8
去分母得：
32=x^2+6x+5
x^2+6x-27=0
(x+9)(x-3)=0
x=-9 or 3