ĐKXĐ: $x\ne -3;1$
$\dfrac{3x-1}{x-1}-1=\dfrac{2x+5}{x+3}-\dfrac{4}{x(x+3)-x-3}\\↔\dfrac{3x-1}{x-1}-1=\dfrac{2x+5}{x+3}-\dfrac{4}{x(x+3)-(x+3)}\\↔\dfrac{(3x-1)(x+3)}{(x-1)(x+3)}-\dfrac{(x-1)(x+3)}{(x-1)(x+3)}=\dfrac{(2x+5)(x-1)}{(x+3)(x-1)}-\dfrac{4}{(x-1)(x+3)}\\→(3x-1)(x+3)-(x-1)(x+3)=(2x+5)(x-1)-4\\↔3x(x+3)-1(x+3)-x(x+3)+1(x+3)=2x(x-1)+5(x-1)-4\\↔3x^2+9x-x-3-x^2-3x+x+3=2x^2-2x+5x-5-4\\↔2x^2+6x=2x^2+3x-9\\↔2x^2+6x-2x^2-3x=-9\\↔3x=-9\\↔x=-3(KTM)$
Vậy pt vô nghiệm $S=\varnothing$
