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Vì (2x+1)^2 \(\ge\) 0, (3x-2y)^2 \(\ge\) 0 \(\Rightarrow\) (2x+1)^2 + (3x-2y) + 2005  \(\ge\)  2005

Vậy A có GTNN là 2005

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................

Chúc bạn học tốt!!!

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 \(\left\{{}\begin{matrix}x=3\\y=\dfrac{3}{2}\end{matrix}\right.\)

\(A=3y^2+2y+5=3\left(y+\frac{1}{3}\right)^2+\frac{14}{3}\ge\frac{14}{3}\)

MIN A = 14/3 khi y = -1/3

\(B=-5x^2+2x-3=-5\left(x-\frac{1}{5}\right)^2-\frac{14}{5}\le-\frac{14}{5}\)

MAX B = -14/5 khi x = 1/5

dùng bđt bunhiacopski thôi

hoặc pt \(\left(x+2y\right)^2=\left(x\cdot1+\sqrt{2}y\cdot\sqrt{2}\right)^2\)