Question

tìm min A= x^2+5y^2-4xy+2x-8y+202

Answer 1

  \(A=x^2+5y^2-4xy+2x-8y+202\)

    \(=x^2+4y^2+1-4xy-4y+2x+\left(y^2-4y+4\right)+197\)

    \(=\left(x-2y+1\right)^2+\left(y-2\right)^2+197\ge197\forall x;y\)

Dâu "=" xảy ra khi: 

\(\hept{\begin{cases}x-2y+1=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}x-4+1=0\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}}\)

Vậy min A = 197 khi \(x=3,y=2\)

Chúc bạn học tốt.