B=\(2x^2-4xy-2x+4y^2+2013\)

\(=x^2-4xy+4y^2+x^2-2x+1+2012\)

\(=\left(x-2y\right)^2+\left(x-1\right)^2+2012\ge2012\)

Dấu = xảy ra khi : \(\left(x-1\right)^2=0\Leftrightarrow x=1\)

                              \(\left(x-2y\right)^2=0\Leftrightarrow2y=1\Leftrightarrow y=\dfrac{1}{2}\)

Vậy \(Min_B=2012\) khi x=1 , y=\(\dfrac{1}{2}\)

\(x^2+y^2-x+4y+5\)

\(=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+4y+4\right)+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\left(y+2\right)^2+\frac{3}{4}\)

\(\ge\frac{3}{4}\)

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

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

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

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

\(\ge-\frac{13}{4}\)

Dấu "=" xảy ra khi \(x=\frac{3}{2};y=-\frac{3}{4}\)

I don't know

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

Sorry !

\(A=x-x^2=-\left(x^2-2\times x\times\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)

\(\left(x-\frac{1}{2}\right)^2\ge0\)

\(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\le\frac{1}{4}\)

Vậy Max A = \(\frac{1}{4}\) khi x = \(\frac{1}{2}\)

***

\(B=5-8x-x^2=-\left(x^2+2\times x\times4+4^2-4^2-5\right)=-\left[\left(x+4\right)^2-21\right]\)

\(\left(x+4\right)^2\ge0\)

\(\left(x+4\right)^2-21\ge-21\)

\(-\left[\left(x+4\right)^2-21\right]\le21\)

Vậy Max B = 21 khi x = - 4 

***

\(C=5-x^2+2x-4y^2-4y=-\left(x^2-2\times x\times1+1^2-1^2+\left(2y\right)^2-2\times2y\times1+1^2-1^2-5\right)=-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\)

\(\left(x-1\right)^2\ge0\)

\(\left(2y-1\right)^2\ge0\)

\(\left(x-1\right)^2+\left(2y-1\right)^2-7\ge-7\)

\(-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\le7\)

Vậy Max C = 7 khi x = 1 và y = \(\frac{1}{2}\)

\(P=8x^2+2y^2+4xy-2x+4y+2015=2\cdot\left(y^2+2xy+2y+4x^2-x\right)+2015\)

\(=2\cdot\left(y^2+2y\left(x+1\right)+\left(x+1\right)^2-\left(x+1\right)^2+4x^2-x\right)+2015\)

\(=2\cdot\left[\left(y+\left(x+1\right)\right)^2+3x^2-3x-1\right]+2015\)

\(=2\cdot\left[\left(y+x+1\right)^2+3\left(x^2-2x\cdot\frac{1}{2}+\frac{1}{4}\right)-1-\frac{3}{4}\right]+2015\)

\(=2\cdot\left[\left(y+x+1\right)^2+3\cdot\left(x-\frac{1}{2}\right)^2\right]+2015-\frac{7}{2}\)

\(=2\cdot\left(x+y+1\right)^2+6\left(x-\frac{1}{2}\right)^2+2011\frac{1}{2}\)

Vậy GTNN của P = 2011,5. Xảy ra khi x=0,5 và y=-1,5.

Nãy lộn nhé,em làm lại:

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

\(=\left[x^2+2x\left(2y+1\right)+\left(2y+1\right)^2\right]+x^2+8\)

\(=\left(x+2y+1\right)^2+x^2+8\ge8\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x^2=0\\x+2y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-\frac{1}{2}\end{cases}}\)

Dạng này mình không quen cho lắm nên không chắc nha!

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

\(=\left[x^2+2x\left(2y+1\right)+\left(2y+1\right)\right]+8\)

\(=\left(x+2y+1\right)^2+8\ge8\)

Dấu "=" xảy ra khi \(\left(x+2y+1\right)^2=0\Leftrightarrow2y+1=-x\)

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

\(=x^2-2x^2-x=-x^2-x=0\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Thay vào D loại x = -1 suy ra x = 0 tức là y = -1/2

\(\text{x}^2+y^2-\text{x}+4y+5=\left(\text{x}^2-\text{x}+\frac{1}{4}\right)+\left(y^2+4y+4\right)+\frac{3}{4}=\left(\text{x}-\frac{1}{2}\right)^2+\left(y+2\right)^2+\frac{3}{4}\) 

\(\ge0+0+\frac{3}{4}=\frac{3}{4}\).Dâu"=" xayr ra khi: 

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