\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x+3\right)-x\left(x-2\right)-\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\left(ĐK:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x^2+6x+x+3-x^2+2x-x^2+9}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow9x+9=0\)
\(\Leftrightarrow9x=-9\)
\(\Leftrightarrow x=-1\left(n\right)\)
Vậy \(S=\left\{-1\right\}\)