Đáp án:

$\begin{array}{l}
Đkxđ:x \ne 2\\
\left( {x - 1} \right)\left( {4 - x} \right) + \frac{{{x^3} + {x^2} - 6x - 2}}{{x - 2}} = 13\\
 \Rightarrow \left( { - {x^2} + 5x - 4} \right)\left( {x - 2} \right) + {x^3} + {x^2} - 6x - 2 = 13\left( {x - 2} \right)\\
 \Rightarrow  - {x^3} + 2{x^2} + 5{x^2} - 10x - 4x + 8 + {x^3} + {x^2} - 6x - 2 = 13x - 26\\
 \Rightarrow 8{x^2} - 33x + 32 = 0\\
 \Rightarrow x = \frac{{33 \pm \sqrt {65} }}{{16}}\left( {tmdk} \right)
\end{array}$