\(\left\{{}\begin{matrix}\left|x-2013\right|\ge0\\\left|1007-\dfrac{1}{2}y\right|\ge0\end{matrix}\right.\Rightarrow\left|x-2013\right|+\left|1007-\dfrac{1}{2}y\right|\ge0\)
Mà \(\left|x-2013\right|+\left|1007-\dfrac{1}{2}y\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-2013\right|=0\\\left|1007-\dfrac{1}{2}y\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2013=0\\1007-\dfrac{1}{2}y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2013\\y=2014\end{matrix}\right.\)
Vậy \(x=2013;y=2014\)
\(\left\{{}\begin{matrix}\left|x-2013\right|\ge0\forall x\in R\\\left|1007-\dfrac{1}{2}y\right|\ge0\forall y\in R\end{matrix}\right.\) đẳng thức chi khi trong dấu trị tuyệt đối =0 " 1/2y --> y ở tử hay mẫu ?"
\(\Rightarrow\left|x-2013\right|+\left|1007-\dfrac{1}{2y}\left(y\right)\right|=0\Rightarrow\) \(\left\{{}\begin{matrix}x-2013=0\\007-\dfrac{1}{2y}\left(y\right)=0\end{matrix}\right.\)
