Điều kiện $\color{red}{x\neq 0,-1,-2,-3,-4,-5,-6,-7}$
Pt $\Leftrightarrow ( \frac{1}{x}+\frac{1}{x+7})+(\frac{1}{x+2}+\frac{1}{x+5})=(\frac{1}{x+1}+\frac{1}{x+6})+(\frac{1}{x+3}+\frac{1}{x+4})$$\Leftrightarrow\frac{2x+7}{x(x+7)}+\frac{2x+7}{(x+2)(x+5)}-\frac{2x+7}{(x+1)(x+6)}-\frac{2x+7}{(x+3)(x+4)}=0$
$\Leftrightarrow(2x+7)[(\frac{1}{x^2+7}-\frac{1}{x^2+7x+6})+(\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+12)]}=0$
Dễ thấy $\frac{1}{x^2+7}>\frac{1}{x^2+7x+6},\frac{1}{x^2+7x+10}>\frac{1}{x^2+7x+12}$
$\Rightarrow(\frac{1}{x^2+7}-\frac{1}{x^2+7x+6})+(\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+12)}\neq 0$
$\Rightarrow2x+7=0\Rightarrow \color{red}{x=\frac{-7}{2}}$ (thõa đk)