C = ( x² - 4xy + 4y² ) + 10.(x -2y) + ( y² -2y + 1) + 27

   = ( x-2y)² + 2.5.(x-2y) + 25 + (y-1)² + 2

   = ( x-2y + 5 )² + (y-1)² + 2 \(\ge2\)vì \(\left(x-2y+5\right)^2\ge0\forall x,y\) và \(\left(y-1\right)^2\ge0\forall y\)

Dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy Min C = 2 \(\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

C = x2 - 4xy + 5y2 + 10x - 22y + 28

= (x^2 - 4xy + 4y^2) + (10x - 20y) + (y^2 - 2y) + 28

= (x - 2y)^2 + 10(x - 2y) + 25 + (y^2 - 2y + 1) + 2

= (x - 2y)^2 + 2.(x - 2y).5 + 5^2 + (y - 1)^2 + 2

= (x - 2y + 5)^2 + (y - 1)2 + 2

Vì (x−2y+5)^2≥0∀x;y; (y−1)^2≥0∀y nên (x−2y+5)^2+(y−1)^2+2≥2∀x;y

hay C≥2∀x;y

Dấu ''='' xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2y-5\\y=1\end{cases}\Rightarrow}\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)

Ta có: 5x2+10y2-6xy-4x-2y +3= x2 -6xy +(3y)2 +4x2 +y2 -4x -2y +3

= (x - 3y)2 +(2x)2 -4x+1+ y2 -2y+1 +1

= (x-3y)2 + (2x -1)2 + (y-1)2 +1

Ta có :(x-3y)2 luôn lớn hơn hoặc bằng 0

(2x -1)2 luôn lớn hơn hoặc bằng 0

(y-1)2 luôn lớn hơn hoặc bằng 0

=>(x-3y)2 + (2x -1)2 + (y-1)2 luôn lớn hơn hoặc bằng 0

=>(x-3y)2 + (2x -1)2 + (y-1)2 +1 >0

ta có:\(A=x^2+5y^2-4xy-2y+2x+2010\)

\(=x^2+4y^2+y^2-4xy-4y+2y+2x+1+1+2008\)

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

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

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

Vì: (x-2y+1)²+(y+1)>0 với \(\forall x;y\)

do đó: (x-2y+1)²+(y+1)+2008 > 2008 với \(\forall x;y\)

Dấu "=" xảy ra khi x-2y+1=0 và y+1=0

ta có:

y+1=0=>y=0-1=>y=-1

thay y=-1 và x-2y+1=0

=>x-2.(-1)+1=0

=>x+2+1=0

=>x+2=-1

=>x=-1-2

=>x=-3

vậy \(A_{min}=2008\) khi x=-3 hoặc x=-1

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

\(R=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+28\)

\(R=\left[\left(x-2y\right)^2+2\left(x-2y\right).5+25\right]+\left(y^2-2y+1\right)+2\)

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

Mà  \(\left(x-2y+5\right)^2\ge0\forall x;y\)

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

\(\Rightarrow R\ge2\)

Dấu "=" xảy ra khi : 

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy ...

\(VT=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\left(x-y\right)^2\left(x+y\right)^2=VP\)

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

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

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

\(A=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)

\(A_{min}=3\) khi \(x=-2\)

\(B=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)

\(B_{min}=1\) khi \(x=10\)

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

\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

\(C_{min}=2\) khi \(\left(x;y\right)=\left(-3;1\right)\)