\(\left\{{}\begin{matrix}\left|x\right|+\dfrac{\sqrt{y}}{\sqrt{y}-2}=5\\\left|x\right|-\dfrac{1}{\sqrt{y}-2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x\right|=5-\dfrac{\sqrt{y}}{\sqrt{y}-2}\\\left|x\right|=1+\dfrac{1}{\sqrt{y}-2}\end{matrix}\right.\)(đk: \(y\ge0,y\ne4\))
\(\Rightarrow5-\dfrac{\sqrt{y}}{\sqrt{y}-2}=1+\dfrac{1}{\sqrt{y}-2}\)
\(\Rightarrow\dfrac{1+\sqrt{y}}{\sqrt{y}-2}=4\)
\(\Rightarrow4\left(\sqrt{y}-2\right)=1+\sqrt{y}\)
\(\Rightarrow4\sqrt{y}-\sqrt{y}=1+8\)
\(\Rightarrow3\sqrt{y}=9\Rightarrow\sqrt{y}=3\Rightarrow y=9\)(thoả mãn đk)
Ta có: \(\left|x\right|=1+\dfrac{1}{\sqrt{9}-2}\)
\(\Rightarrow\left|x\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy nghiệm (x;y) \(\in\left\{\left(2;9\right),\left(-2;9\right)\right\}\)
