ĐK : $x,y\geq 0$
Xét (x,y)=(0,0) không là ngiệm của hệ nên ta có $\left\{\begin{matrix}
1+\frac{1}{x+y}=\frac{2}{\sqrt{3x}} & \\ 1-\frac{1}{x+y}=\frac{4\sqrt{2}}{\sqrt{7y}}
&
\end{matrix}\right.$
$\left\{\begin{matrix}
1=\frac{1}{\sqrt{3x}}+\frac{2\sqrt{2}}{\sqrt{7y}} & \\\frac{1}{x+y}=\frac{1}{\sqrt{3x}}-\frac{2\sqrt{2}}{\sqrt{7y}}
&
\end{matrix}\right.$
Nhân theo vế ta có $\frac{1}{x+y}=\begin{pmatrix} \frac{1}{\sqrt{3x}} \end{pmatrix}^{2}-\begin{pmatrix} \frac{2\sqrt{2}}{\sqrt{7y}} \end{pmatrix}^{2}=\frac{1}{3x}-\frac{8}{7y}\Leftrightarrow (4x+7y)(6x-y)=0$ mà $x,y> 0\Rightarrow 4x+7y> 0\Rightarrow 6x-y=0\Leftrightarrow x=\frac{y}{6}\Rightarrow 1=\frac{\sqrt{2}}{\sqrt{y}}+\frac{2\sqrt{2}}{\sqrt{7y}}\Leftrightarrow 1=\frac{2\sqrt{14}+7\sqrt{2}}{7\sqrt{y}}\Leftrightarrow y=\frac{22+8\sqrt{7}}{7}\Rightarrow x=\frac{11+4\sqrt{7}}{21}$
Edited by NGOCTIEN_A1_DQH, 18-08-2012 - 10:24.