`x/(x-3)-x/(x-5)=x/(x-4)-x/(x-6)`
`<=>x (\frac{1}{x-3}-\frac{1}{x-5})=x(\frac{1}{x-4}-\frac{1}{x-6})`$\left[\begin{array}{l}x=0\\\frac{1}{x-3}-\frac{1}{x-5}=\frac{1}{x-4}-\frac{1}{x-6}\end{array} \right.$
$⇔ \frac{(x-5)-(x-3)}{(x-3)(x-5)}=\frac{(x-6)-(x-4)}{(x-4)(x-6)}$
$⇔ \frac{-2}{(x-3)(x-5)}=\frac{-2}{(x-4)(x-6)}$
$⇔ (x-3)(x-5)=(x-4)(x-6)$
`⇔x^2−8x+15=x^2−10x+24`
`⇔2x=9`
`⇔x=9/2`
Vậy `S={0;9/2}`
