Giúp mik 1 câu này thui thực hiện phép tính á

#andy

\[\begin{array}{l}
\dfrac{{{x^2} - 4x - 9}}{{2{x^2} - 6x}} + \dfrac{{x - 5}}{{3 - x}} + \dfrac{{ - 3}}{{2x}}\\
 = \dfrac{{{x^2} - 4x - 9}}{{2x\left( {x - 3} \right)}} - \dfrac{{x - 5}}{{x - 3}} - \dfrac{3}{{2x}}\\
DKXD:\left\{ \begin{array}{l}
2x \ne 0\\
x - 3 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne 3
\end{array} \right.\\
MC:2x\left( {x - 3} \right)\\
 = \dfrac{{{x^2} - 4x - 9}}{{2x\left( {x - 3} \right)}} - \dfrac{{\left( {x - 5} \right)2x}}{{2x\left( {x - 3} \right)}} - \dfrac{{3\left( {x - 3} \right)}}{{2x\left( {x - 3} \right)}}\\
 = \dfrac{{{x^2} - 4x - 9}}{{2x\left( {x - 3} \right)}} - \dfrac{{2{x^2} - 10x}}{{2x\left( {x - 3} \right)}} - \dfrac{{3x - 9}}{{2x\left( {x - 3} \right)}}\\
 = \dfrac{{{x^2} - 4x - 9 - 2{x^2} + 10x - 3x + 9}}{{2x\left( {x - 3} \right)}}\\
 = \dfrac{{ - {x^2} + 3x}}{{2x\left( {x - 3} \right)}}\\
 = \dfrac{{ - x\left( {x - 3} \right)}}{{2x\left( {x - 3} \right)}}\\
 =  - \dfrac{1}{2}
\end{array}\]