a.
\(\left\{{}\begin{matrix}\left|x-1\right|+\left|y-2\right|=2\\\left|x-1\right|+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y-2\right|-y=-1\\\left|x-1\right|+y=3\end{matrix}\right.\)
Xét phương trình: \(\left|y-2\right|-y=-1\)
TH1: \(y\ge2\)
\(\Rightarrow y-2-y=-1\Leftrightarrow-2=-1\) (loại)
TH2: \(y\le2\)
\(\Rightarrow2-y-y=-1\Rightarrow y=\dfrac{3}{2}\)
Thế vào \(\left|x-1\right|+y=3\)
\(\Rightarrow\left|x-1\right|+\dfrac{3}{2}=3\Rightarrow\left|x-1\right|=\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{3}{2}\Rightarrow x=\dfrac{5}{2}\\x-1=-\dfrac{3}{2}\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}\left|x+1\right|+\left|y-1\right|=5\\\left|x+1\right|-4y+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y-1\right|+4y-4=5\\\left|x+1\right|-4y+4=0\end{matrix}\right.\)
Xét phương trình: \(\left|y-1\right|+4y-4=5\)
TH1: \(y\ge1\)
\(\Rightarrow y-1+4y-4=5\Rightarrow y=2\)
Thế vào \(\left|x+1\right|-4y+4=0\)
\(\Rightarrow\left|x+1\right|=4\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
TH2: \(y\le1\)
\(\Rightarrow1-y+4y-4=5\Rightarrow y=\dfrac{8}{3}>1\) (không thỏa mãn)
