a, \(A=x^2+2y^2+2xy+2x-4y+2013\)
\(=\left(x^2+y^2+2xy\right)+\left(2x+2y\right)+\left(y^2-6y+9\right)+2004\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1+\left(y-3\right)^2+2003\)
\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2003\ge2003\forall x,y\in R\)
Dấu = xảy ra khi: \(\left\{{}\begin{matrix}x+y+1=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3+1=0\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)
Vậy GTNN của A = 2003 khi x = -4, y = 3.
b, \(B=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)\)
\(=\left(x^2-2x-5x+10\right)\left(x^2-7x-10\right)\)
\(=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)
\(=\left(x^2-7x\right)^2-10^2\)
\(=\left(x^2-7x\right)^2-100\ge-100\forall x\in R\)
Dấu = xảy ra khi: \(x^2-7x=0\Leftrightarrow x\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
Vậy GTNN của B = -100 khi x = 0 hoặc x = 7.
c, \(C=x^2+y^2-xy-x+y+1\)
\(4C=4x^2+4y^2-4xy-4x+4y+4\)
\(4C=\left(4x^2-4xy+y^2\right)-\left(4x-2y\right)+3y^2+2y+4\)
\(4C=\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+3\)
\(4C=\left(2x-y-1\right)^2+3y^2+2y+3\)
\(12C=3\left(2x-y-1\right)^2+9y^2+6y+9\)
\(12C=3\left(2x-y-1\right)^2+\left(9y^2+6y+1\right)+8\)
\(12C=3\left(2x-y-1\right)^2+\left(3y+1\right)^2+8\)
\(\Rightarrow12C\ge8\Leftrightarrow C\ge\dfrac{2}{3}\forall x,y\in R\)
Dấu = xảy ra khi: \(\left\{{}\begin{matrix}2x-y-1=0\\3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{1}{3}-1=0\\y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy GTNN của C = \(\dfrac{2}{3}\) khi \(x=\dfrac{1}{3},y=-\dfrac{1}{3}\).
d, \(D=x^2-2xy+2y^2-4y+5\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+1\)
\(=\left(x-y\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\in R\)
Dấu = xảy ra khi: \(\left\{{}\begin{matrix}x-y=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy GTNN của D = 1 khi x = 2, y = 2.
