Question

tìm gtnn của C:C=x^2+xy+y^2+x+y+202

Answer 1

\(C=\left(x^2+\dfrac{y^2}{4}+\dfrac{1}{4}+xy+x+\dfrac{y}{2}\right)+\dfrac{1}{12}\left(9y^2+6y+1\right)+\dfrac{605}{3}\)

\(C=\left(x+\dfrac{y}{2}+\dfrac{1}{2}\right)^2+\dfrac{1}{12}\left(3y+1\right)^2+\dfrac{605}{3}\ge\dfrac{605}{3}\)

\(C_{min}=\dfrac{605}{3}\) khi \(x=y=-\dfrac{1}{3}\)