\(D=x^2+20y^2+8xy-4y+2009\)

\(\Leftrightarrow D=x^2+16y^2+4y^2+8xy-4y+1+2008\)

\(\Leftrightarrow D=\left(x^2+8xy+16y^2\right)+\left(4y^2-4y+1\right)+2008\)

\(\Leftrightarrow D=\left[x^2+2.x.4y+\left(4y\right)^2\right]+\left[\left(2y\right)^2-2.2y.1+1^2\right]+2008\)

\(\Leftrightarrow D=\left(x+4y\right)^2+\left(2y-1\right)^2+2008\)

Vậy GTNN của \(D=2008\) khi \(\left\{{}\begin{matrix}x+4y=0\\2y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+4.\left(0,5\right)=0\\y=0,5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=0,5\end{matrix}\right.\)

a) \(C=x^2-4xy+5y^2+10x-22y+28\)

\(\Leftrightarrow C=x^2-4xy+4y^2+y^2+10x-20y-2y+1+25+2\)

\(\Leftrightarrow C=\left(x^2-4xy+4y^2\right)+\left(10x-20y\right)+\left(y^2-2y+1\right)+2+25\)

\(\Leftrightarrow C=\left(x-2y\right)^2+10\left(x-2y\right)+\left(y-1\right)^2+2+25\)

\(\Leftrightarrow C=\left[\left(x-2y\right)^2+10\left(x-2y\right)+25\right]+\left(y-1\right)^2+2\)

\(\Leftrightarrow C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\)

Vậy GTNN của \(C=2\) khi \(\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-2.1+5=0\\y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

P=4x²+4xy+y²+x²-4x+4+y²+8y+16+5

=> P=(2x+y)²+ (x-2)² + (y+4)² +5

Ta nhận thấy: \(\hept{\begin{cases}\left(2x+y\right)^2\ge0\forall x,y\\\left(x-2\right)^2\ge0\forall x\\\left(y+4\right)^2\ge0\forall y\end{cases}}\)

=> P=(2x+y)²+ (x-2)² + (y+4)² +5 \(\ge\)5  Với mọi x, y

=> GTNN của P là P_min = 5

Đạt được khi: 

\(\hept{\begin{cases}\left(2x+y\right)^2=0\\\left(x-2\right)^2=0\\\left(y+4\right)^2=0\end{cases}}\) <=> \(\hept{\begin{cases}2x+y=0\\x-2=0\\y+4=0\end{cases}}\) <=> \(\hept{\begin{cases}x=2&y=-4&\end{cases}}\)

\(A=4x^2-4xy+5y^2+20x-6y+2044\)

\(=\left(4x^2-4xy+y^2\right)+20x-6y+4y^2+2044\)

\(=\left(2x-y\right)^2+10\left(2x-y\right)+25+\left(4y^2+4y+1\right)+2018\)

\(=\left(2x-y+5\right)^2+\left(2y+1\right)^2+2018\ge2018\)

Dấu "=" xảy ra tại \(y=-\frac{1}{2};x=-\frac{11}{4}\)

Ta có \(A=4x^2-4xy+5y^2+20x-6y+2044\)

            \(=4x^2-4x\left(y-5\right)+\left(y-5\right)^2+4y^2+4y+1+2018\)

            \(=\left(2x-y+5\right)^2+\left(2y+1\right)^2+2018\)

Vì...\(\Rightarrow A\ge2018\)

Dấu bằng xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-y+5=0\\2y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{11}{4}\\y=-\frac{1}{2}\end{cases}}}\)

Mấy bạn giải chi tiết ra giùm mình

\(C=2x^2+4y^2+4xy-3x-1\)

\(=\left(x^2+4xy+4y^2\right)+\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{13}{4}\)

\(=\left(x+2y\right)^2+\left(x-\dfrac{3}{2}\right)^2-\dfrac{13}{4}\)

Ta có : \(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x-\dfrac{3}{2}\right)^2\ge0\end{matrix}\right.\) \(\Leftrightarrow P\ge-\dfrac{13}{4}\)

Dấu "=" xảy ra khi :

\(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\\left(x-\dfrac{3}{2}\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-\dfrac{3}{4}\end{matrix}\right.\)

Vậy \(C_{Min}=-\dfrac{13}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-\dfrac{3}{4}\end{matrix}\right.\)

C= 2x² +4y²+4xy -3x -1

Mk viết nhầm đề các bạn thông cảm nhé

\(E=5x^2+y^2+10+4xy-14x-6y\)

\(E=\left(2x+y-3\right)^2+\left(x-1\right)^2+6\)

Vì \(\left(2x+y-3\right)^2+\left(x-1\right)^2\ge0\)

Dấu '=" xảy ra.......................