a,Ta có : \(\left\{{}\begin{matrix}\left|x-3\right|\ge0\\\left|2x-y+1\right|\ge0\end{matrix}\right.\\ \Rightarrow\left|x-3\right|+\left|2x-y+1\right|\ge0\\ \Leftrightarrow\left|x-3\right|+\left|2x-y+1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\2x-y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
Vậy x = 3 ; y = 7
b, \(\left\{{}\begin{matrix}\left|x-2y+1\right|\ge0\\\left|x-y-2\right|\ge0\end{matrix}\right.\\ \Rightarrow\left|x-2y+1\right|+\left|x-y-2\right|\ge0\\ \Leftrightarrow\left|x-2y+1\right|+\left|x-y-2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2y+1=0\\x-y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\y=3\end{matrix}\right.\)
Vậy x = 5; y = 3
