Đặt  \(A=x^2+y^2+xy+3x+3y+2018\)

\(4.A=4x^2+4y^2+4xy+12x+12y+8072\)

\(4.A=\left(4x^2+4xy+y^2\right)+3y^2+12x+12y+8072\)

\(4.A=\left[\left(2x+y\right)^2+2\left(2x+y\right).3+9\right]+3\left(y^2+2y+1\right)+8060\)

\(4.A=\left(2x+y+3\right)^2+3\left(y+1\right)^2+8060\)

Mà  \(\left(2x+y+3\right)^2\ge0\forall x;y\)

       \(\left(y+1\right)^2\ge0\forall y\)\(\Rightarrow3\left(y+1\right)^2\ge0\forall y\)

\(\Rightarrow4.A\ge8060\)

\(\Leftrightarrow A\ge2015\)

Dấu "=" xảy ra khi : 

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

Vậy ...

Lời giải:

$2Q=2x^2+2xy+2y^2-6x-6y+3998$

$=(x^2+2xy+y^2)+x^2+y^2-6x-6y+3998$

$=(x+y)^2-4(x+y)+(x^2-2x)+(y^2-2y)+3998$

$=(x+y)^2-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)+3992$

$=(x+y-2)^2+(x-1)^2+(y-1)^2+3992\geq 3992$

$\Rightarrow Q\geq 1996$

Vậy $Q_{\min}=1996$ khi $x+y-2=x-1=y-1=0\Leftrightarrow x=y=1$

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$R=(x^2+2xy+y^2)+x^2-2x+2y+15$

$=(x+y)^2+2(x+y)+x^2-4x+15$

$=(x+y)^2+2(x+y)+1+(x^2-4x+4)+10$

$=(x+y+1)^2+(x-2)^2+10\geq 10$
Vậy $R_{\min}=10$ khi $x+y+1=x-2=0$

$\Leftrightarrow x=2; y=-3$

đặt x+y=a; xy=b; ta có \(b\le\frac{a^2}{4}\)

B = \(a^2-b-3a+2019\ge a^2-\frac{a^2}{4}-3a+2019=\frac{3}{4}\left(a-2\right)^2+2016\)\(\ge2016\)

B đạt GTNN khi a= \(2;a^2=4b\) <=> x=y = 1

Bạn nhân 4 lên rồi tách ra hằng đẳng thức

Ta có 

A=x²+xy+y²-3x-3y+2016

=>4A=4x²+4xy+y² -6(2x+y) + 9 + 3(y²-2y+1) +8052

         =(2x+y)²-6(2x+y)+9 + 3(y-1)² +8052 

        =(2x+y-3)²+3(y-1)²+8052>= 8052

     =>A>=2013

Dấu bang xay ra khi x=y=1

Ta có A= x²+xy+y²+3x-3y+2016

=> 2A= 2x²+2xy+2y²+6x-6y+4032

=> 2A=(x²+2xy+y²)+(x²+6x+9)+(y²-6y+9)+ 4014

=> 2A= (x+y)²+ (x+3)²+(y-3)²+4014

=> 2A >= 4014=> A>=2007

Dấu "=" xảy ra khi x=-3; y=-3