\(\Leftrightarrow A=\left(4x^2+y^2+4xy+4x+2y+1\right)+\left(6x^2-2x+\frac{2}{36}\right)+\frac{17}{18}\)
\(\Leftrightarrow A=\left(2x+y+1\right)^2+2\left(\sqrt{3}x+\frac{1}{2\sqrt{3}}\right)^2+\frac{17}{18}\ge\frac{17}{18}\)
Dấu bằng xảy ra khi 2x+y=-1, \(\sqrt{3}x=-\frac{1}{2\sqrt{3}}\)
Giải x,y típ nhé