Vì \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\forall x\\\left(2y-3\right)^2\ge0\forall y\end{matrix}\right.\)\(\Rightarrow\left(x-3\right)^2+\left(2y-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+\left(2y-3\right)^2+2014\ge2014\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(2y-3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(D_{MIN}=2014\) khi \(\left\{{}\begin{matrix}x=3\\y=\dfrac{3}{2}\end{matrix}\right.\)
