a)đặt A=\(x^2+5y^2-2xy+4y+3\)

 \(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)

=\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\)

ta thấy GTNN của A =2 khi x=y=-1/2

\(D=x^2+5y^2+2xy-2y+2005\)

\(D=\left(x^2+2xy+y^2\right)+\left(4y^2-2y+\frac{1}{4}\right)+2004,75\)

\(D=\left(x+y\right)^2+\left(2y+\frac{1}{2}\right)^2+2004,75\)

Mà  \(\left(x+y\right)^2\ge0\forall x;y\)

      \(\left(2y+\frac{1}{2}\right)^2\ge0\forall y\)

\(\Rightarrow D\ge2004,75\)

Dấu "=" xảy ra khi : 

\(\hept{\begin{cases}x+y=0\\2y+\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=-\frac{1}{4}\end{cases}}\)

Vậy  \(D_{Min}=2004,75\Leftrightarrow\left(x;y\right)=\left(\frac{1}{4};-\frac{1}{4}\right)\)

\(D=x^2+5y^2-2xy+4y+3\)

\(=x^2-2xy+y^2+4y^2+4y+1+2\)

\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(2y+1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{1}{2}\)

Vậy \(D_{min}=2\Leftrightarrow x=y=-\dfrac{1}{2}\)

a) Ta có: \(A=x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

c) Ta có: \(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu '=' xảy ra khi x(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

d) Ta có: \(x^2+5y^2-2xy+4y+3\)

\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)