\(A=3x^2-2x+6=\left(3x^2-2\cdot\sqrt{3}x\cdot\dfrac{1}{\sqrt{3}}+\dfrac{1}{3}\right)+\dfrac{17}{3}=\left(\sqrt{3}x-\dfrac{1}{\sqrt{3}}\right)^2+\dfrac{17}{3}\)
Vì: \(\left(\sqrt{3}x-\dfrac{1}{\sqrt{3}}\right)^2\ge0\Rightarrow\left(\sqrt{3}x-\dfrac{1}{\sqrt{3}}\right)^2+\dfrac{17}{3}\ge\dfrac{17}{3}\)
dấu ''='' xảy ra khi x = \(\dfrac{1}{3}\)
vậy MinA \(=\dfrac{17}{3}\Leftrightarrow x=\dfrac{1}{3}\)
-B là tìm gtln nha bn
\(B=-x^2+3x-2=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}\)
vì: \(-\left(x-\dfrac{3}{2}\right)^2\le0\forall x\Rightarrow-\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
dấu ''='' xảy ra khi \(x=\dfrac{3}{2}\)
vậy MaxB = \(\dfrac{1}{4}\Leftrightarrow x=\dfrac{3}{2}\)
