Điều kiện xác định $x\ne -1;-2;-3;-4$
$\dfrac{1}{(x+1)(x+2)}+\dfrac{1}{(x+2)(x+3)}+\dfrac{1}{(x+3)(x+4)}=\dfrac{1}{6}$
$\Rightarrow \dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{1}{6}$
$\Rightarrow \dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}$
$\Rightarrow \dfrac{3}{(x+1)(x+4)}=\dfrac{1}{6}$
$\Rightarrow 18=(x+1)(x+4)$
$\Rightarrow x^2+5x-14=0$
$\Rightarrow (x+7)(x-2)=0\Rightarrow \left[ \begin{array}{l}x=2\\x=-7\end{array} \right.$
