\(A=\left(x^2+x+\frac{1}{4}\right)+\left(y^2+y+\frac{1}{4}\right)-\frac{1}{2}+2019\)
\(A=\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}\right)+\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)+\frac{4037}{2}\)
\(=\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{4037}{2}\ge\frac{4037}{2}\)
Đẳng thức xảy ra khi \(x=y=-\frac{1}{2}\)
\(x^2+y^2+x+y+2019=x^2+y^2+x+y+\frac{1}{4}+\frac{1}{4}+2019-\frac{1}{2}=\left(x^2+x+\frac{1}{4}\right)+\left(y^2+y+\frac{1}{4}\right)+\frac{4037}{2}=\left(x^2+2.\frac{1}{2}x+\frac{1^2}{2^2}\right)+\left(y^2+2.\frac{1}{2}y+\frac{1^2}{2^2}\right)+\frac{4037}{2}=\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{4037}{2}\) \(Vì:\left\{{}\begin{matrix}\left(x+\frac{1}{2}\right)^2\ge0\\\left(y+\frac{1}{2}\right)^2\ge0\end{matrix}\right.nên:A\ge0+0+\frac{4037}{2}=\frac{4037}{2}\)
\(Vậy:A_{min}=\frac{4037}{2}.\)Dâus "=" xay ra khi:\(x=y=-\frac{1}{2}\)
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