\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{x^2+5x+6}-\frac{2}{x-3}=0\)
<=> \(\frac{\left(13-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{6\left(x^2+1\right)}{\left(x-3\right)\left(x+3\right)\left(x^2+1\right)}-\frac{3\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x+3\right)\left(x-3\right)}-\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\)
<=> \(13x-39-x^2+3x+6-3x+9-2x-6=0\)
<=> 11x - 30 - x2 = 0
<=> 5x - 30 + 6x - x2 = 0
<=> 5(x - 6) - x(x - 6) = 0
<=> (5 - x)(x - 6) = 0
<=> \(\left[\begin{matrix}5-x=0\\x-6=0\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=5\\x=6\end{matrix}\right.\)
Vậy S = {5; 6}
=> 5 + 6 = 11
ĐS: 11
