`S=3x^2+2x+1`
`=(3x^2+2x+1/3)+2/3`
`=[(\sqrt3 x)^2+ 2.\sqrt3 x . 1/\sqrt3 + (1/\sqrt3)^2]+2/3`
`=(\sqrt3 x+1/\sqrt3)^2 + 2/3`
`=(\sqrt3x+\sqrt3/3)^2+2/3`
`=> D_(min) =2/3 <=> \sqrt3x+\sqrt3/3=0 <=>x=-1/3`
\(D=3\left(x^2+\dfrac{2}{3}x+\dfrac{1}{3}\right)=3\left(x^2+\dfrac{2}{3}x+\dfrac{1}{9}+\dfrac{2}{9}\right)=3\left(x+\dfrac{1}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\)
dấu"=" xảy ra<=>x=-1/3
Ta có:D=3x2+2x+1
= \(3\left(x^2+2.\dfrac{1}{3}x+\dfrac{1}{9}\right)-\dfrac{1}{3}\)
= \(3\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{3}\)
Vì \(3\left(x-\dfrac{1}{3}\right)^2\ge0\forall x\)
\(\Rightarrow D\ge-\dfrac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Ta có: \(D=3x^2+2x+1\)
\(=3\left(x^2+\dfrac{2}{3}x+\dfrac{1}{3}\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(x+\dfrac{1}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{3}\)
