\(A=xy\left(x-2\right)\left(y+6\right)+12x^2-24x+3y^2+18y+2047\)

   \(=xy\left(x-2\right)\left(y+6\right)+12\left(x^2-2x\right)+3y\left(y+6\right)+2047\)

   \(=y\left(y+6\right)\left(x^2-2x\right)+12\left(x^2-2x+3\right)+3y\left(y+6\right)+2011\)

   \(=y\left(y+6\right)\left(x^2-2x+3\right)+12\left(x^2-2x+3\right)+2011\)

   \(=\left(x^2-2x+3\right)\left(y^2+6y+12\right)+2011\)

   \(=\left[\left(x-1\right)^2+2\right].\left[\left(y+3\right)^2+3\right]+2011\ge2.3+2011=2017\)

Dấu "=" xảy ra khi: 

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

Vậy GTNN của A là 2017 khi \(x=1,y=-3\)

Ta có : 

\(B=x\left(x-2\right)y\left(y+6\right)+12x^2-24x+3y^2+18y+36\)

\(=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y+12\right)+12\)

\(=\left(x^2-2x\right)\left(y^2+6y+12\right)+3\left(y^2+6y+12\right)+12\)

\(=\left(x^2-2x+3\right)\left(y^2+6y+12\right)+12\)

\(=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+12\ge2.3+12=18\)

\(xy\left(x-2\right)\left(y+6\right)+12x^2-24x+3y^2+18y+2045.\)

\(=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y\right)+2045\)

\(=\left[\left(x^2-2x\right)\left(y^2+6y\right)+3\left(y^2+6y\right)\right]+12\left(x^2-2x+3\right)+2009.\)

\(=\left(x^2-2x+3\right)\left(y^2+6x\right)+12\left(x^2-2x+3\right)+2009\)

\(=\left(x^2-2x+3\right)\left(y^2+6x+12\right)+2009\)

\(=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+2009\)

Ta có: \(\left(x-1\right)^2\ge0\forall x\Leftrightarrow\left(x-1\right)^2+2\ge2\)

\(\left(y+3\right)^2\ge0\forall y\Leftrightarrow\left(y+3\right)^2+3\ge3\)

Suy ra \(B=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+2009\ge2.3+2009=2015\)

Vậy GTNN của B=2015 khi x=1, y=-3.

\(P=xy\left(x-2\right)\left(y+6\right)+12x^2-24x+3y^2+18y+36\)

\(=\left(x^2-2x\right)\left(y^2+6y\right)+\left(12x^2+24x+12\right)+\left(3y^2+18y+9\right)+15\)

\(=\left[\left(x-1\right)^2-1\right]\left[\left(y+3\right)^2-9\right]+12\left(x-1\right)^2+3\left(y+3\right)^2+15\)

\(=3\left(x-1\right)^2+2\left(y+3\right)^2+15\)

Do đó \(P\ge15\)

\(\Rightarrow P>0\)

Suy ra P luôn dương

\(P=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y\right)+36\)

\(=\left(x^2-2x\right)\left(y^2+6y+12\right)+3\left(y^2+6y+12\right)\)

\(=\left(x^2-2x+3\right)\left(y^2+6y+12\right)\)

\(=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]>0\)

\(P=xy\left(x-2\right)\left(x+6\right)+12x^2-24x+3y^2+18y+36\)

\(=xy\left(x-2\right)\left(x+6\right)+12x\left(x-2\right)+3y\left(y+6\right)+36\)

Đặt \(\left\{{}\begin{matrix}x-2=a\\x+6=b\end{matrix}\right.\) . Khi đó

\(P=xy.a.b+12x.a+3y.b+36\)

Phân tích tiếp ....