题目内容：

1/(x2+x)+1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)=4/21
x2是x的平方

优质解答

1/(x2+x)+1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)
=1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/x-1/(x+4)
=4/(x²+4x)=4/21
x²+4x-21=0
(x+7)(x-3)=0
x1=-7
x2=3
检验得x1=-7,x2=3是原方程的解