Đáp án: $x = \frac{{9 \pm \sqrt {73} }}{2}$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x \ne 3;x \ne - 3\\
\frac{x}{{x + 3}} - \frac{{x - 2}}{{2x - 6}} = \frac{{x + 2}}{{{x^2} - 9}}\\
\Rightarrow \frac{x}{{x + 3}} - \frac{{x - 2}}{{2\left( {x - 3} \right)}} = \frac{{x + 2}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
\Rightarrow \frac{{x.2\left( {x - 3} \right) - \left( {x - 2} \right)\left( {x + 3} \right)}}{{2\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{2x + 4}}{{2\left( {x - 3} \right)\left( {x + 3} \right)}}\\
\Rightarrow 2{x^2} - 6x - {x^2} - x + 6 = 2x + 4\\
\Rightarrow {x^2} - 9x + 2 = 0\\
\Rightarrow {\left( {x - \frac{9}{2}} \right)^2} = \frac{{73}}{4}\\
\Rightarrow x = \frac{{9 \pm \sqrt {73} }}{2}\left( {tmdk} \right)
\end{array}$
