$\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\\ \Leftrightarrow \dfrac{1}{x^2+2x+3x+6}+\dfrac{1}{x^2+3x+4x+12}+\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}=\dfrac{1}{8}\\ \Leftrightarrow \dfrac{1}{(x+2)(x+3)}+\dfrac{1}{(x+3)(x+4)}+\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}=\dfrac{1}{8}\\ \Leftrightarrow \dfrac{x+3-(x+2)}{(x+2)(x+3)}+\dfrac{x+4-(x+3)}{(x+3)(x+4)}+\dfrac{x+5-(x+4)}{(x+4)(x+5)}+\dfrac{x+6-(x+5)}{(x+5)(x+6)}=\dfrac{1}{8}\\ \Leftrightarrow \dfrac{x+3}{(x+2)(x+3)}-\dfrac{x+2}{(x+2)(x+3)}+\dfrac{x+4}{(x+3)(x+4)}-\dfrac{x+3}{(x+3)(x+4)}+\dfrac{x+5}{(x+4)(x+5)}-\dfrac{x+4}{(x+4)(x+5)}+\dfrac{x+6}{(x+5)(x+6)}-\dfrac{x+5}{(x+5)(x+6)}=\dfrac{1}{8}\\ \Leftrightarrow \dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\\ \Leftrightarrow \dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\\ \Leftrightarrow \dfrac{x+6-(x+2)}{(x+2)(x+6)}=\dfrac{1}{8}\\ \Leftrightarrow \dfrac{4}{(x+2)(x+6)}=\dfrac{1}{8}\\ \Leftrightarrow (x+2)(x+6)=32\\ \Leftrightarrow x^2+8x−20=0\\ \Leftrightarrow (x-2)(x+10)=0\\ \Leftrightarrow \left[\begin{array}{l} x=2\\ x=-10\end{array} \right.$
