Question

Tìm GTNN: A=x²+y²-2y+12x+40

Answer 1

\(A=\left(x^2+12x+36\right)+\left(y^2-2y+1\right)+3\\ A=\left(x+6\right)^2+\left(y-1\right)^2+3\ge3\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=1\end{matrix}\right.\)

Answer 2

\(A=x^2+y^2-2y+12x+40\)

\(=x^2+12x+36+y^2-2y+1+3\)

\(=\left(x+6\right)^2+\left(y-1\right)^2+3\ge3\forall x,y\)

Dấu '=' xảy ra khi x=-6 và y=1