Vì \(\left\{{}\begin{matrix}\left|x^2+2x\right|\ge0\\\left|y^2-y\right|\ge0\end{matrix}\right.\)mà \(\left|x^2+2x\right|+\left|y^2-y\right|=0\Rightarrow\left\{{}\begin{matrix}x^2+2x=0\\y^2-y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\y\left(y-1\right)=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\\\left[{}\begin{matrix}y=0\\y-1=0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\\left[{}\begin{matrix}y=0\\y=1\end{matrix}\right.\end{matrix}\right.\)
Lời giải:
Ta thấy:
$|x^2+2x|\geq 0$ với mọi $x$
$|y^2-y|\geq 0$ với mọi $y$
Do đó để $|x^2+2x|+|y^2-2y|=0$ thì $x^2+2x=y^2-2y=0$
\(\Rightarrow \left\{\begin{matrix} x(x+2)=0\\ y(y-2)=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} \left[\begin{matrix} x=0\\ x=-2\end{matrix}\right.\\ \left[\begin{matrix} y=0\\ y=2\end{matrix}\right.\end{matrix}\right.\)
Vậy $(x,y)=(0,0); (0,2); (-2,0); (-2,2)$
