$\frac{5x-3}{x^2-9}$-$\frac{x}{x-3}$=$\frac{2x-1}{x+3}$ Đkxđ: x$\neq$±3
⇔$\frac{5x-3}{(x+3)(x-3)}$-$\frac{x(x+3)}{(x+3)(x-3)}$=$\frac{(2x-1)(x-3)}{(x+3)(x-3)}$
⇔5x-3-x²-3x=2x²-6x-x+3
⇔-3x²+9x-6=0
⇔-[(3x²-3x)-(6x-6)]=0
⇔-3(x-1)(x-2)=0
⇔\(\left[ \begin{array}{l}x-1=0\\x-2=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=1(thỏa.mãn.đkxđ)\\x=2(thỏa.mãn.đkxđ)\end{array} \right.\)
Vậy S={1;2}.
