$\dfrac{x+3}{x-3} =$ $\dfrac{3}{x^2-3x}+$ $\dfrac{1}{x}$ ⇔ $\dfrac{x+3}{x-3} =$ $\dfrac{3}{x(x-3)}+$ $\dfrac{1}{x}$
Điều kiện $x$$\neq$$0; x$$\neq$$3$
$⇔$ $\dfrac{x(x+3)}{x(x-3)}=$ $\dfrac{3}{x(x-3)}+$ $\dfrac{1(x-3)}{x(x-3)}$
$⇔ x(x+3)=3+(x-3) ⇔ x²+3x=3+x-3$
$⇔ x²+3x-3-x+3=0$
$⇔ x²+2x=0$
$⇔ x(x+2)=0$
$⇔ x=0$ (loại) hoặc $x=-2$ (nhận)
Vậy $S=${$-2$}
BẠN THAM KHẢO NHA!!!
