Đáp án:
$ĐKXĐ: x\ne 8;9;10;11$
\[\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\]
\[\to \dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\]
\[\to \dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10} \]
\[\to x\times \left(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}\right)=0\] \[\to \left[ \begin{array}{l}x=0 ( TMDKXD)\\\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\to \dfrac{1}{x-9}-\dfrac{1}{x-8}=\dfrac{1}{x-11}-\dfrac{1}{x-10}\to x=\dfrac{19}{2} ( TMDKXD) \end{array} \right.\]
