A = \(x^2-3x+1\)
\(=x^2-2x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{5}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}\)
Vì : \(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)
=> A\(\ge-\dfrac{5}{4}\)
Dấu "=" xảy ra khi : \(\left(x-\dfrac{3}{2}\right)^2=0\)
=> x = \(\dfrac{3}{2}\)
Vậy Min A = \(\dfrac{-5}{4}\) khi \(x=\dfrac{3}{2}\)
b) B = \(9x^2+x-1\)
\(=\left(3x\right)^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{5}{4}\)
\(=\left(3x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}\)
Vì \(\left(3x+\dfrac{1}{2}\right)^2\ge0\forall x\)
nên : \(\left(3x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)
=> B \(\ge\dfrac{-5}{4}\)
Dấu "=" xảy ra khi : \(\left(3x+\dfrac{1}{2}\right)^2=0\)
=> 3x = \(\dfrac{-1}{2}\) => x = \(\dfrac{-1}{6}\)
Vậy Min B = \(\dfrac{-5}{4}\) khi x = \(\dfrac{-1}{6}\)
