\(\left\{{}\begin{matrix}\frac{1}{\left|x-2\right|}+y=3\\\frac{2}{\left|x-2\right|}-\frac{y}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{\left|x-2\right|}=3-y\\2\left(3-y\right)-\frac{y}{2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{\left|x-2\right|}=3-y\\-2y-\frac{y}{2}=1-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{\left|x-2\right|}=3-y\\-2,5y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{\left|x-2\right|}=3-2=1\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1=\left|x-2\right|\\y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\veebar x=1\\y=2\end{matrix}\right.\)
Vậy giải hệ PT ta được x = 3 hoặc x = 1 và y = 2