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 ...

Đặt P =\(x^2+xy+y^2-3x-3y+2018\)

= \(x^2+\left(xy-3x\right)+y^2-3y+2018\)

= \(x^2+x\left(y-3\right)+y^2-3y+2018\)

= \(x^2+2.x.\dfrac{y-3}{2}+\dfrac{\left(y-3\right)^2}{4}-\dfrac{\left(y-3\right)^2}{4}+y^2-3y+2018\)

= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{-y^2+6y-9+4y^2-12y}{4}+2018\)

= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3y^2-6y-9}{4}+2011\)

= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3}{4}\left(y^2-2y-3\right)+2018\)

\(=\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2+2015\)

Với mọi x;y có \(\left(x+\dfrac{y-3}{2}\right)^2\ge0\) ; \(\dfrac{3}{4}\left(y-1\right)^2\ge0\)

\(\Rightarrow\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2+2015\ge2015\) với mọi x;y

\(\Rightarrow P\ge2015\) với mọi x;y

\(P=2015\Leftrightarrow\) \(\left\{{}\begin{matrix}x+\dfrac{y-3}{2}=0\\y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+y-3=0\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy ......

Ta có: \(G=x^2+xy+y^2-3x-3y\)

\(=\left(x^2+2xy+y^2\right)-3\left(x+y\right)-xy\)

\(=\left(x+y\right)^2-3\left(x+y\right)-xy\)

Mà \(\left(x+y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\)

\(\Leftrightarrow x^2+2xy+y^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow xy\le\frac{\left(x+y\right)^2}{4}\Leftrightarrow-xy\ge-\frac{\left(x+y\right)^2}{4}\)

\(\Rightarrow G\ge\frac{\left(x+y\right)^2-3\left(x+y\right)-\left(x+y\right)^2}{4}\)

\(\Leftrightarrow G\ge\frac{3\left(x+y\right)^2}{4}-3\left(x+y\right)\)

Đến đây để cho dễ nhìn, ta đặt \(t=x+y\)

\(\Rightarrow G\ge\frac{3t^2}{4}-3t=3\left(\frac{t^2}{4}-\frac{2t}{2}+1\right)-3\ge3\left(\frac{t}{2}-1\right)^2-3\ge3\)

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

Vậy \(MIN_G=-3\Leftrightarrow x=y=1\)

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 $x+y+1=x-2=0$

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