Vijetnam 2007

The necessary conditions for the given equation system are $x,y>0;y+3x\ne 0$. The given equation system is equivalent to the following equation systems $$\{\begin{array}{l}1-\frac{12}{y+3x}=\frac{2}{\sqrt{x}}\\ 1+\frac{12}{y+3x}=\frac{6}{\sqrt{x}}\end{array}\Rightarrow \{\begin{array}{l}\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=1\\ -\frac{1}{\sqrt{x}}+\frac{3}{\sqrt{y}}=\frac{12}{y+3x}\end{array}(1)(2)$$ Multiply (1) with (2), we get $$\frac{9}{y}-\frac{1}{x}=\frac{12}{y+3x}\iff y^2+6xy-27x^2=0\iff \{\begin{array}{l}y=3x\\ y=-9x\end{array}$$ By the condition $x,y>0$, we have $y=3x$. By setting $y=3x$ in (1), we obtain $$\frac{1}{\sqrt{x}}+\frac{3}{\sqrt{3x}}=1\Rightarrow x=4+2\sqrt{3},y=12+6\sqrt{3}$$