\(A=2x^2+6x\\ =2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
Với mọi x;y thì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\\ \Rightarrow2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Hay \(A\ge-\dfrac{9}{2}\)
Để \(A=\dfrac{9}{2}\) thì \(\left(x+\dfrac{3}{2}\right)^2=0\)
=>\(x+\dfrac{3}{2}=0\)
=>\(x=-\dfrac{3}{2}\)
Vậy...