Bình phương 2 vế ( 2 vế ko âm ), ta được:$2\sqrt{3}-3==x\sqrt{3}-2\sqrt{3xy}+y\sqrt{3}$$\Leftrightarrow \sqrt{3}(2-x-y)=3-2\sqrt{3xy}$$\Leftrightarrow 3(2-x-y)^2=9+12xy-12\sqrt{3xy}$$\Leftrightarrow \sqrt{3xy}=\frac{9+12xy-3(2-x-y)^2}{12}$$\rightarrow 3-2\sqrt{3xy}\epsilon Q$$\rightarrow \sqrt{3}(2-x-y)\epsilon Q$ mà $\sqrt{3}\epsilon I$$\Rightarrow 2-x-y=0$$\Rightarrow \left\{ \begin{array}{l} x+y=2\\ 2\sqrt{3xy}=3 \end{array} \right.$Giải hệ được: $(x;y)\epsilon {(\frac{1}{2};\frac{3}{2});(\frac{3}{2};\frac{1}{2})}$

ĐK$:\left\{ \begin{array}{l} x,y\geq0 \\ x\geq y \end{array} \right.$Bình phương 2 vế ( 2 vế ko âm ), ta được:$2\sqrt{3}-3==x\sqrt{3}-2\sqrt{3xy}+y\sqrt{3}$$\Leftrightarrow \sqrt{3}(2-x-y)=3-2\sqrt{3xy}$$\Leftrightarrow 3(2-x-y)^2=9+12xy-12\sqrt{3xy}$$\Leftrightarrow \sqrt{3xy}=\frac{9+12xy-3(2-x-y)^2}{12}$$\rightarrow 3-2\sqrt{3xy}\epsilon Q$$\rightarrow \sqrt{3}(2-x-y)\epsilon Q$ mà $\sqrt{3}\epsilon I$$\Rightarrow 2-x-y=0$$\Rightarrow \left\{ \begin{array}{l} x+y=2\\ 2\sqrt{3xy}=3 \end{array} \right.$Giải hệ được: $(x;y)\epsilon {(\frac{1}{2};\frac{3}{2});(\frac{3}{2};\frac{1}{2})}$Đối chiếu đk đc:.............