\(\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{x^2-2x-3}\)
* x2 - 2x - 3 = x2- 3x + x - 3 = x(x-3 ) + ( x - 3) = ( x - 3 ) ( x + 1 )
\(\Leftrightarrow\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm3;x\ne-1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)+8\left(x+3\right)=2x\left(x+3\right)\)
\(\Leftrightarrow x^2-2x+1+8x+24=2x^2+6x\)
\(\Leftrightarrow-x^2+25=0\)
\(\Leftrightarrow x^2-25=0\Leftrightarrow\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vậy \(S=\left\{-5;5\right\}\)
