Ta có
$\frac{x}{x-3}$- $\frac{x}{x-5}$ =$\frac{x}{x-4}$ -$\frac{x}{x-6}$ (ĐKXĐ:x khác 3,4,5,6)
⇔$\frac{1}{x-3}$.x- $x$.$\frac{1}{x-5}$= $\frac{1}{x-4}$.x- $\frac{1}{x-6}$.x
⇔x.($\frac{1}{x-3}$-$\frac{1}{x-5}$)=x.($\frac{1}{x-4}$-$\frac{1}{x-6}$)
⇔x.$\frac{x-5-x+3}{x^2-8x+15}$=x.$\frac{x-6-x+4}{x^2-10x+24 }$
⇔$\frac{-2x}{x^2-8x+15}$=$\frac{-2x}{x^2-10x+24}$
⇔$\left \{ {{x^2-8x+15-x^2+10x-24=0} \atop {-2x=0}} \right.$
⇔$\left \{ {{2x-8=0} \atop {x=0}} \right.$
⇔$\left \{ {{x=4(không thõa mãn)} \atop {x=0}} \right.$
Vậy tập nghiệm S là S={0}
