Question

- Tìm Min
a. A=4x^2 +8x +y^2-4y+20
b. B=2x^2+2xy +5y^2-8x-22y

Answer 1

\(A=4x^2+8x+y^2-4y+20\)

\(A=\left(4x^2+8x\right)+\left(y^2-4y\right)+20\)

\(A=4\left(x^2+2x+1\right)+\left(y^2-4y+4\right)-4-4+20\)

\(A=4\left(x+1\right)^2+\left(y-2\right)^2+12\ge12\forall x,y\)

Do \(4\left(x+1\right)^2\ge0\forall x;\left(y-2\right)^2\ge0\forall y\)

Dấu "=" Xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy Min A=12 <=>\(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)