Đáp án:
$\begin{array}{l}
Dkxd:x \ne 2;x \ne 4\\
\dfrac{2}{{ - {x^2} + 6x - 8}} - \dfrac{{x - 1}}{{x - 2}} = \dfrac{{x + 3}}{{x - 4}}\\
\Rightarrow \dfrac{2}{{{x^2} - 6x + 8}} + \dfrac{{x - 1}}{{x - 2}} + \dfrac{{x + 3}}{{x - 4}} = 0\\
\Rightarrow \dfrac{{2 + \left( {x - 1} \right)\left( {x - 4} \right) + \left( {x + 3} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 4} \right)}} = 0\\
\Rightarrow 2 + {x^2} - 5x + 4 + {x^2} + x - 6 = 0\\
\Rightarrow 2{x^2} - 4x = 0\\
\Rightarrow 2x\left( {x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\left( {tm} \right)\\
x = 2\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 0
\end{array}$
