ta có :
\(\left\{{}\begin{matrix}\left|x-1\right|+\left|y-2\right|=2\\\left|x-1\right|+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-\left|y-2\right|=1\\\left|x-1\right|+y=3\end{matrix}\right.\)
ta có
\(\left|y-2\right|=\left\{{}\begin{matrix}y-2nếuy\ge2\\2-ynếuy< 2\end{matrix}\right.\)
Nếu \(y\ge2\) ta có hệ pt:
\(\left\{{}\begin{matrix}y-\left(y-2\right)=1\\\left|x-1\right|+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-y+2=1\\\left|x-1\right|+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2=1\\\left|x-1\right|+y=3\end{matrix}\right.\) (vô lí)
\(\Rightarrow y\ge2\) (loại )
Nếu y< 2 ta có hệ pt:
\(\left\{{}\begin{matrix}y-\left(2-y\right)=1\\\left|x-1\right|+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2+y=1\\\left|x-1\right|+y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y-2=1\\\left|x-1\right|+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{2}\\\left|x-1\right|+\frac{3}{2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{2}\\\left|x-1\right|=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{2}\\\left[{}\begin{matrix}x-1=\frac{3}{2}\\x-1=\frac{-3}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{2}\left(TM\right)\\\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-1}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\frac{5}{2}\\y=\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\frac{-1}{2}\\y=\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
