\(A=2x^2-3x\)
\(=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}-\dfrac{9}{16}\right)\)
\(=2\left[\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{16}\right]\)
\(=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)
Vậy Min A là : \(-\dfrac{9}{8}\Leftrightarrow x=\dfrac{3}{4}\)
A=2x2-3x
2A =4x2-6x
= 4x2-2.2.\(\dfrac{6}{4}\).x+\(\dfrac{36}{16}-\dfrac{36}{16}\)
=(2x-\(\dfrac{6}{4}\))2-\(\dfrac{36}{16}\)
ta có \(\left(2x-\dfrac{6}{4}\right)^2\ge0\Rightarrow\left(2x-\dfrac{6}{4}\right)^4-\dfrac{36}{16}\ge\dfrac{-36}{16}\)
=> 2A đạt GTNN = \(\dfrac{-36}{16}\) tại x=\(\dfrac{6}{8}\)
=> A đạt GTNN =\(\dfrac{-13}{16}\) tại x =\(\dfrac{6}{8}\)