Question

Tìm giá trị nhỏ nhất : \(A=x^2-2xy+6y^2-12x+2y+45\)

Answer 1

\(A=x^2-2xy+6y^2-12x+2y+45\)

\(=\left(x^2-2xy+y^2-12x+12y+36\right)+\left(5y^2-10y+5\right)+4\)

\(=\left[\left(x-y\right)^2-12\left(x+y\right)+6^2\right]+5\left(y^2-2y+1\right)+4\)

\(=\left(x-y+6\right)^2+5\left(y-1\right)^2+4\)

Ta có: \(\left(x-y+6\right)^2\ge0\forall x,y\)

\(5\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-y+6\right)^2+5\left(y-1\right)^2+4\ge4\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow x=7,y=1\)

Vậy \(A_{MIN}=4\Leftrightarrow x=7,y=1\)