题目

1/x²+x+1/x²+3x+2+1/x²+5x+6+1/x²+7x+12,分式方程,

提问时间：2020-07-26

答案

原式=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)