Đáp án: $x = \dfrac{{ \pm \sqrt {57} - 3}}{2}$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x \ne 3;x \ne - 3\\
\dfrac{x}{{x + 3}} - \dfrac{{x - 2}}{{2x + 6x}} = \dfrac{{3 - 2x}}{{{x^2} - 9}}\\
\Rightarrow \dfrac{{2x - x + 2}}{{2\left( {x + 3} \right)}} = \dfrac{{3 - 2x}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
\Rightarrow \dfrac{{x + 2}}{{2\left( {x + 3} \right)}} = \dfrac{{3 - 2x}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
\Rightarrow \dfrac{{\left( {x + 2} \right).\left( {x - 3} \right)}}{{2\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{2.\left( {3 - 2x} \right)}}{{2.\left( {x - 3} \right)\left( {x + 3} \right)}}\\
\Rightarrow {x^2} - x - 6 = 6 - 4x\\
\Rightarrow {x^2} + 3x - 12 = 0\\
\Rightarrow x = \dfrac{{ \pm \sqrt {57} - 3}}{2}\left( {tmdk} \right)
\end{array}$
