\(\frac{x_1\left(x_2-1\right)+x_2\left(x_1-1\right)}{\left(x_1-1\right)\left(x_2-1\right)}=\frac{13}{6}\Leftrightarrow\frac{2x_1x_2-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}=\frac{13}{6}\)
\(\Leftrightarrow\frac{2x_1x_2-1}{x_1x_2}=\frac{13}{6}\Leftrightarrow12x_1x_2-6=13x_1x_2\Rightarrow x_1x_2=-6\)
Theo Viet đảo, \(x_1;x_2\) là nghiệm:
\(x^2-x-6=0\)