$\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}$
$\to \dfrac{8}{x-8}+\dfrac{11}{x-11}-\dfrac{9}{x-9}-\dfrac{10}{x-10}=0$
$\to (\dfrac{8}{x-8}+1)+(\dfrac{11}{x-11}+1)-(\dfrac{9}{x-9}+1)-(\dfrac{10}{x-10}+1)=0$
$\to \dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}-\dfrac{9+x-9}{x-9}-\dfrac{10+x-10}{x-10}=0$
$\to \dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0$
$\to x(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10})=0$
Ta có:
$\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{x-10}=0 $
$\to \dfrac{1}{x-8}+\dfrac{1}{x-11}=\dfrac{1}{x-9}+\dfrac{1}{x-10}$
$\to \dfrac{x-8+x-11}{(x-8)(x-11)}=\dfrac{x-9+x-10}{(x-9)(x-10)}$
$\to \dfrac{2x-19}{(x-8)(x-11)}=\dfrac{2x-19}{(x-9)(x+10)}$
$\to \dfrac{2x-10}{(x-8)(x-11)}-\dfrac{2x-19}{(x-10)(x-9)}=0$
$\to (2x-19)(\dfrac{1}{(x-8)(x-11)}-\dfrac{1}{(x-9)(x-10)})=0$
$\to 2x-19=0 \to x=\dfrac{19}{2}$
$x=0 \to x=0$
Vậy $S={0;\dfrac{19}{2}}$
