Ta có : \(D=3x^2+2x+1=3\left(x^2+\frac{2}{3}x+\frac{1}{3}\right)=3\left(x^2+\frac{2}{3}x+\frac{1}{9}+\frac{2}{9}\right)=3\left(x+\frac{1}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)
\(\Rightarrow\)Min D = 2/3
Dấu "=" xảy ra khi x + 1/3 = 0
\(\Rightarrow x=-\frac{1}{3}\)
Vậy Min D = 2/3 khi x = -1/3
D = 3x2 + 2x + 1 = 3( x2 + 2/3x + 1/9 ) + 2/3 = 3( x + 1/3 )2 + 2/3 ≥ 2/3 ∀ x
Dấu "=" xảy ra <=> x = -1/3 . Vậy MinD = 2/3