a) Dễ thấy |2x+1| \(\ge\) 0 và \(\left|y-1\right|\ge0\)
Lại có: \(\left|2x+1\right|+\left|y-1\right|=4\)
mà |2x+1| \(⋮̸\) 2 \(\Rightarrow\) \(\left|2x+1\right|+\left|y-1\right|=4=1+3=3+1\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|2x+1\right|=1;\left|y-1\right|=3\\\left|2x+1\right|=3;\left|y-1\right|=1\end{matrix}\right.\)
+ Với: |2x+1|=1 ; |y-1| =3, ta có:
\(2x+1\in\left\{1;-1\right\}\) \(\Leftrightarrow x\in\left\{0;-1\right\}\)
y-1\(\in\left\{3;-3\right\}\) \(\Leftrightarrow y\in\left\{4;-2\right\}\)
+ Với \(\left|2x+1\right|\)=3 ; |y-1|=1 ; ta có:
\(2x+1\in\left\{3;-3\right\}\) \(\Leftrightarrow x\in\left\{1;-2\right\}\)
\(y-1\in\left\{1;-1\right\}\) \(\Leftrightarrow y\in\left\{2;0\right\}\)
Vậy: \(\left(x;y\right)\in\left\{\left(0;4\right);\left(0;-2\right);\left(-1;4\right);\left(-1;-2\right);\left(1;2\right);\left(1;0\right);\left(-2;2\right);\left(-2;0\right)\right\}\)
