\(2x^2-4xy+8y^2+7x+6y-15.\)
= \(x^2+x^2-4xy+4y^2+4y^2+7x+6y-15\)
= \(\left(x^2-4xy+4y^2\right)+\left[x^2+7x+\left(\frac{7}{2}\right)^2\right]+\left[4y^2+6y+\left(\frac{3}{2}\right)^2\right]-\left(\frac{7}{2}\right)^2-\left(\frac{3}{2}\right)^2-15\)
= \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2-\frac{59}{2}\)
Vì \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2\ge0\forall x;y\)
=> \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2-\frac{59}{2}\ge0-\frac{59}{2}\forall x;y\)
=> \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2-\frac{59}{2}\ge-\frac{59}{2}\)
Vậy GTNN của bt là \(\frac{-59}{2}\Leftrightarrow\hept{\begin{cases}x-2y=0\\x+\frac{7}{2}=0\\2y+\frac{3}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2y\Rightarrow\orbr{\begin{cases}x=-\frac{7}{4}\\y=-\frac{3}{2}\end{cases}}\\x=-\frac{7}{2}\\y=-\frac{3}{4}\end{cases}}\)
