Đáp án:
\(x = \pm \sqrt {\frac{{18051}}{{2003}}} \)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \pm 3\\
\frac{3}{{x - 3}} - \frac{6}{{9 - {x^2}}} + \frac{x}{{x + 3}} = 2004\\
\to \frac{{3x + 9 + 6 + {x^2} - 3x}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{{2004\left( {{x^2} - 9} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
\to {x^2} + 15 = 2004{x^2} - 9.2004\\
\to 2003{x^2} - 18051 = 0\\
\to {x^2} = \frac{{18051}}{{2003}}\\
\to x = \pm \sqrt {\frac{{18051}}{{2003}}} \left( {TM} \right)
\end{array}\)
