\(B=5x^2+9y^2-12xy+24x-48y+81\)
\(B=4x^2-12xy+9y^2+16\left(2x-3y\right)+64+x^2-8x+16+1\)
\(B=\left(2x-3y\right)^2+2\left(2x-3y\right).8+8^2+\left(x-4\right)^2+1\)
\(B=\left(2x-3y+8\right)^2+\left(x-4\right)^2+1\)
ta có :\(\left(2x-3y+8\right)^2\ge0\)
\(\left(x-4\right)^2\ge0\)
\(\Leftrightarrow\left(2x-3y+8\right)^2+\left(x-4\right)^2+1\ge1\)
\(\Rightarrow B\ge1\Rightarrow\)GTNN của B là 1
dấu"=" xảy ra khi \(\left\{{}\begin{matrix}2x-3y+8=0\\x-4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+8=3y\\x=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{16}{3}\\x=4\end{matrix}\right.\)
