A = 2x2 +2x+1 = 2(x2 +x+\(\dfrac{1}{2}\) ) = 2(x2 + x+\(\dfrac{1}{4}\) ) -\(\dfrac{1}{2}\)
=2(x+\(\dfrac{1}{2}\) )2 -\(\dfrac{1}{2}\)
\(A=2x^2+2x+1=2\left(x^2+x+\dfrac{1}{2}\right)=2\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{4}\right]=2\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\right]=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)
Vì: \(2\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)
Dấu ''='' xảy ra khi \(x=-\dfrac{1}{2}\)
Vậy \(MIN_A=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)
