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

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

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

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

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

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

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.

Ta có : 

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

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

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

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

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

x²-2x+1=(x-1)² >= 0 => x²-2x+3 >= 2 với mọi x thuộc R (1)

y²+6y+9=(y+3)² >=0 => y²+6y+12 >=3 với mọi y thuộc R (2)

M=xy(x-2)(y+6)-12x²-24x+3y²+18y+2050

=(x²-2x)(y²+6y)+12(x²-2x)+3(y²+6y)+36+2014

=(x²-2x)(y²+6y+12)+3(y²+6y+12)+2014

=(x²-2x+3)(y²+6y+12)+2014 (3)

từ (1); (2) và (3) => B >= 2.3+2014 => B >= 2020

dấu "=" xảy ra <=> x=1 và y=-3

vậy minM=2020 khi x=1; y=-3