Đáp án:
$\begin{array}{l}
Dkxd:x \ne 3;4;5;6\\
\dfrac{x}{{x - 3}} + \dfrac{x}{{x - 6}} = \dfrac{x}{{x - 4}} + \dfrac{x}{{x - 5}}\\
\Rightarrow x.\left( {\dfrac{1}{{x - 3}} + \dfrac{1}{{x - 6}} - \dfrac{1}{{x - 4}} - \dfrac{1}{{x - 5}}} \right) = 0\\
\Rightarrow x.\left( {\dfrac{{x - 6 + x - 3}}{{\left( {x - 3} \right)\left( {x - 6} \right)}} - \dfrac{{x - 5 + x - 4}}{{\left( {x - 4} \right)\left( {x - 5} \right)}}} \right) = 0\\
\Rightarrow x.\left( {\dfrac{{2x - 9}}{{\left( {x - 3} \right)\left( {x - 6} \right)}} - \dfrac{{2x - 9}}{{\left( {x - 4} \right)\left( {x - 5} \right)}}} \right) = 0\\
\Rightarrow x.\left( {2x - 9} \right).\left( {\dfrac{1}{{\left( {x - 3} \right)\left( {x - 6} \right)}} - \dfrac{1}{{\left( {x - 4} \right)\left( {x - 5} \right)}}} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
2x - 9 = 0\\
\dfrac{1}{{\left( {x - 3} \right)\left( {x - 6} \right)}} = \dfrac{1}{{\left( {x - 4} \right)\left( {x - 5} \right)}}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{9}{2}\\
{x^2} - 9x + 18 = {x^2} - 9x + 20\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x \in \left\{ {0;\dfrac{9}{2}} \right\}
\end{array}$
