ĐKXĐ: x≠8
Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)
\(\Leftrightarrow\frac{9}{6\left(x-8\right)}+\frac{6\left(3x-20\right)}{6\left(x-8\right)}+\frac{6\left(x-8\right)}{48\left(x-8\right)}-\frac{2\left(13x-102\right)}{6\left(x-8\right)}=0\)
\(\Leftrightarrow9+6\left(3x-20\right)+6\left(x-8\right)-2\left(13x-102\right)=0\)
\(\Leftrightarrow9+18x-120+6x-48-26x+204=0\)
\(\Leftrightarrow45-2x=0\)
\(\Leftrightarrow2x=45\)
hay \(x=\frac{45}{2}\)(tm)
Vậy: \(x=\frac{45}{2}\)