Question

Tìm Min:

A=\(18x^2+12y^2+12x+12y+7\)

Answer 1

\(A=18x^2+12y^2+12x+12y+7\)

\(=2\left(9x^2+6x\right)+3\left(4y^2+4y\right)+7\)

\(=2\left(9x^2+6x+1-1\right)+3\left(4y^2+4y+1-1\right)+7\)

\(=2\left(3x+1\right)^2-2+3\left(2y+1\right)^2-3+7\)

\(=2\left(3x+1\right)^2+3\left(2y+1\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(3x+1\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(MinA=2\)