có $\sqrt{x-1}$-$\sqrt{y-1}$=$y\sqrt{y}$-$x\sqrt{x}$<=>$\frac{(\sqrt{x-1}-\sqrt{y-1})(\sqrt{x-1}+\sqrt{y-1})}{\sqrt{x-1}+\sqrt{y-1}}$=-($\sqrt{x}$-$\sqrt{y}$)(x+$\sqrt{xy}$+y)biến đổi tiếp
có $\sqrt{x-1}$-$\sqrt{y-1}$=$y\sqrt{y}$-$x\sqrt{x}$<=>$\frac{(\sqrt{x-1}-\sqrt{y-1})(\sqrt{x-1}+\sqrt{y-1})}{\sqrt{x-1}+\sqrt{y-1}}$=-($\sqrt{x}$-$\sqrt{y}$)(x+$\sqrt{xy}$+y)<=>$\frac{x-y}{\sqrt{x-1}+\sqrt{y-1}}$+($\sqrt{x}$-$\sqrt{y}$)(x+$\sqrt{xy}$+y)=0<=>($\sqrt{x}$-$\sqrt{y}$)($\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x-1}+\sqrt{y-1}}$+$x$+$\sqrt{xy}$+$y$)=0<=>$\sqrt{x}$=$\sqrt{y}$<=>x=ythay x=y vào pt ta đc...