Với mọi giá trị của \(x;y\in R\) ta có:
\(\left(x-3\right)^2+\left(2y-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+\left(2y-3\right)^2+2014\ge2014\)
Hay \(D\ge2014\) với mọi giá trị của \(x;y\in R\)
Để \(D=2014\) thì \(\left(x-3\right)^2+\left(2y-3\right)^2+2014=2014\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(2y-3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=\dfrac{3}{2}\end{matrix}\right.\)
Vậy................
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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 \(\left\{{}\begin{matrix}x=3\\y=\dfrac{3}{2}\end{matrix}\right.\)
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