đkxđ : y-2> 0 = > y > 2
Đặt : \(\dfrac{1}{\sqrt{y}-2}=a\Leftrightarrow\left\{{}\begin{matrix}x+a=3\\\dfrac{x}{2}-3a=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\a=1\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{\sqrt{y}-2}=1\Leftrightarrow y-2=1\Rightarrow y=3\left(tm\right)\)
\(\Rightarrow\left(x;y\right)=\left(2;3\right)\)