$\dfrac{1}{(x+1)(x+2)}+\dfrac{1}{(x+2)(x+3)}+...+\dfrac{1}{(x+7)(x+8)}=\dfrac{1}{14}$ ($x\ne -1;-2;...;-8$)
$↔\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+7}-\dfrac{1}{x+8}=\dfrac{1}{14}$
$↔\dfrac{1}{x+1}-\dfrac{1}{x+8}=\dfrac{1}{14}$
$↔\dfrac{x+8-x-1}{(x+1)(x+8)}=\dfrac{(x+1)(x+8)}{14(x+1)(x+8)}$
$↔98=x²+9x+8$
$↔0=x²+9x-90$
$↔0=(x-6)(x+15)$
$↔x-6=0\quad or\quad x+15=0$
$↔x=6(TM)\quad or\quad x=-15(TM)$