\(\left(x^2+4x+4\right)+\left(y^2-2.y.\frac{1}{2}+\frac{1}{4}\right)+\frac{8055}{4}\ge\frac{8055}{4}\)
\(x^2+y^2+4x-y+2018\)
\(=x^2+4x+4+y^2-y+\frac{1}{4}+\frac{8055}{4}\)
\(=\left(x+2\right)^2+\left(y-\frac{1}{2}\right)^2+\frac{8055}{4}\ge\frac{8055}{4}\forall x;y\)
Dấu"=" xả ra<=> \(\hept{\begin{cases}\left(x+2\right)^2=0\\\left(y-\frac{1}{2}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\y=\frac{1}{2}\end{cases}}}\)
Vậy.
\(x^2+y^2+4x-y+2018\)
\(=\left(x^2+4x+4\right)+\left(y^2-y+\frac{1}{4}\right)+\frac{8055}{4}\)
\(=\left(x+2\right)^2+\left(y-\frac{1}{2}\right)^2+\frac{8055}{4}\ge\frac{8055}{4}\)
Vậy GTNN của bt là \(\frac{8055}{4}\)
\(\Leftrightarrow\hept{\begin{cases}x+2=0\\y-\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=\frac{1}{2}\end{cases}}\)
