ĐK: $x\geq 1$$2(\sqrt{x-1}-1)+(\sqrt{5x-1}-3)=x^2-4$
$\Rightarrow 2.\frac{x-2}{\sqrt{x-1}+1}+\frac{5x-1-9}{\sqrt{5x-1}+3}=(x+2)(x-2)$
$\Leftrightarrow (x-2)\frac{2}{\sqrt{x-1}+1}+(x-2)\frac{5}{\sqrt{5x-1}+3}=(x+2)(x-2)$
$\Rightarrow x=2$ (nhận) hoặc
$\frac{2}{\sqrt{x-1}+1}+\frac{5}{\sqrt{5x-1}+3}=x+2$
$\Leftrightarrow x+2-3-(\frac{2}{\sqrt{x-1}+1}-2)-(\frac{5}{\sqrt{5x-1}+3}-1)=0$
$\Leftrightarrow (x-1)+\frac{2\sqrt{x-1}}{\sqrt{x-1}+1}+\frac{5(x-1)}{(\sqrt{5x-1}+3).(\sqrt{5x-1}+2)}=0$
$\Rightarrow x=1$( nhận)
vậy $S={1;2}$