D ez nhất :v
\(D=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+5\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+5\ge5\)
Đẳng thức xảy ra khi x = 1 và y = -2
\(A=\left[\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4\right]+\left(y^2-2y+1\right)+2020\)
\(=\left[\left(x-y\right)^2+2\left(x-y\right).2+2^2\right]+\left(y-1\right)^2+2020\)
\(=\left(x-y+2\right)^2+\left(y-1\right)^2+2020\ge2020\)
Dấu "=" xảy ra khi y = 1 và x - y + 2 = 0 tức là x = y - 2 = -1
\(B=\left(x^2-2xy+y^2\right)-2\left(x-y\right)+1+x^2-2x+1+2019\)
\(=\left(x-y\right)^2-2\left(x-y\right).1+1+\left(x-1\right)^2+2019\)
\(=\left(x-y-1\right)^2+\left(x-1\right)^2+2019\ge2019\)
Dấu "=" xảy ra khi x = 1 và x - y - 1 = 0 hay y = 0
\(C=\left(x^2+2.x.2y+4y^2\right)-2.\left(x+2y\right).3+9+x^2-2x+1+2010\)
\(=\left(x+2y\right)^2-2\left(x+2y\right).3+9+\left(x-1\right)^2+2010\)
\(=\left(x+2y-3\right)^2+\left(x-1\right)^2+2010\ge2010\)
Dấu "=" xảy ra khi x = 1 và x + 2y - 3 = 0 hay y = 1
