Ta có: \(\dfrac{6}{x-5}+\dfrac{2}{x-8}=\dfrac{18}{\left(x-5\right)\left(8-x\right)}-1\)
\(\Leftrightarrow6x-48+2x-10=-18-\left(x-5\right)\left(x-8\right)\)
\(\Leftrightarrow8x-58+18+x^2-13x+40=0\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=5\left(loại\right)\end{matrix}\right.\)