\(\begin{cases}2x-|y-2|=2\\x+2|y-2|=6\end{cases}\\↔\begin{cases}2x-|y-2|=2\\2x+4|y-2|=12\end{cases}\\↔\begin{cases}2x-|y-2|=2\\2x-|y-2|+5|y-2|=12\end{cases}\\↔\begin{cases}2x=2+|y-2|\\2+5|y-2|=12\end{cases}\\↔\begin{cases}x=\dfrac{2+|y-2|}{2}\\5|y-2|=10\end{cases}\\↔\begin{cases}x=\dfrac{2+|y-2|}{2}\\|y-2|=2\end{cases}\\↔\begin{cases}x=2\\\left[\begin{array}{1}y-2=2\\y-2=-2\end{array}\right.\end{cases}\\↔\begin{cases}x=2\\\left[\begin{array}{1}y=4\\y=0\end{array}\right.\end{cases}\)
Vậy hệ pt có nghiệm \( (x;y)=(2;4);(2;0)\)
