\(2B=2x^2+2y^2-2xy-6x-6y+4058\)

\(2B=\left(x-y\right)^2+\left(x-3\right)^2+\left(y-3\right)^2+4040\ge4040\)

\(\Rightarrow B\ge2020\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-y=0\\x-3=0\\y-3=0\end{cases}\Leftrightarrow x=y=3}\)

Vậy ....

Lời giải:

Ta có:

$A=x^2+xy+y^2-3x-3y+2008$
$2A=2x^2+2xy+2y^2-6x-6y+4016$

$=(x^2+2xy+y^2)-4(x+y)+4+ (x^2-2x+1)+(y^2-2y+1)+ 4010$

$=(x+y)^2-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)+4010$

$=(x+y-2)^2+(x-1)^2+(y-1)^2+4010\geq 4010$

$\Rightarrow A\geq 2005$

Vậy $A_{\min}=2005$

Giá trị này đạt tại $x+y-2=x-1=y-1=0$

$\Leftrightarrow x=y=1$

A=x^2-2x+y^2-2y-x-y+xy

A+3=x^2-2x+1+y^2-2y+1-x-y+xy+1=(x-1)^2+(y-1)^2+(x-1)(y-1)

dat x-1=a;y-1=b

=>A+3=a^2+b^2+ab =a^2+1/4b^2+ab+3/4b^2=(a+1/2b)^2+3/4b^2

=>A+3>=0 <=>x=1;y=1

=>Amin =-3<=> x=1;y=1

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

\(P=\left(\left(x-1\right)^2-\left(x-1\right)\left(y-1\right)+\frac{\left(y-1\right)^2}{4}\right)+\frac{3\left(y-1\right)^2}{4}+2012=\left(x-1-\frac{y-1}{2}\right)^2+\frac{3\left(y-1\right)^2}{4}+2012\ge2012\)

=> Min P=2012 <=> \(\frac{2x-2-y+1}{2}=0\Leftrightarrow2x-y-1=0\) và \(\frac{3\left(y-1\right)^2}{4}=0\Leftrightarrow y=1\)=> \(2x-1-1=0\Leftrightarrow x=1\)