$\left| x \right|+\left| y \right|=3$
Vì $x\in \mathbb{N},y\in \mathbb{Z}$ nên ta có các trường hợp sau:
$T{{H}_{1}}:$
$\begin{cases}\left|x\right|=0\\\left|y\right|=3\end{cases}$$\Leftrightarrow$$\begin{cases}x=0\\\left[\begin{array}{l}y=3\\y=-3\end{array} \right.\end{cases}$
$T{{H}_{2}}:$
$\begin{cases}\left|x\right|=1\\\left|y\right|=2\end{cases}$$\Leftrightarrow$$\begin{cases}x=1\\\left[ \begin{array}{l}y=2\\y=-2\end{array} \right.\end{cases}$
$T{{H}_{3}}:$
$\begin{cases}\left|x\right|=2\\\left|y\right|=1\end{cases}$$\Leftrightarrow$$\begin{cases}x=2\\\left[ \begin{array}{l}y=1\\y=-1\end{array} \right.\end{cases}$
$T{{H}_{4}}:$
$\begin{cases}\left|x\right|=3\\\left|y\right|=0\end{cases}$$\Leftrightarrow$$\begin{cases}x=3\\y=0\end{cases}$
Vậy có:...
