\(Q=2x-2-3x^2\\ Q=2x-\dfrac{1}{3}-\dfrac{5}{3}-3x^2\\ Q=-\left(3x^2-2x+\dfrac{1}{3}\right)-\dfrac{5}{3}\\ Q=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{5}{3}\\ Q=-3\left[x^2-2\cdot x\cdot\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]-\dfrac{5}{3}\\ Q=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{5}{3}\\ Do\left(x-\dfrac{1}{3}\right)^2\ge0\forall x\\ \Rightarrow-\left(x-\dfrac{1}{3}\right)^2\le0\forall x\\ \Rightarrow-3\left(x-\dfrac{1}{3}\right)^2\le0\forall x\\ \Rightarrow Q=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{5}{3}\le-\dfrac{5}{3}\forall x\)
Dấu \("="\) xảy ra khi :
\(\left(x-\dfrac{1}{3}\right)^2=0\\ \Leftrightarrow x-\dfrac{1}{3}=0\\ \Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(Q_{\left(Max\right)}=-\dfrac{5}{3}\) khi \(x=\dfrac{1}{3}\)
Ta có: \(Q=2x-2-3x^2=-3\left(x^2+\dfrac{2}{3}x+\dfrac{2}{3}\right)\)
\(=-3\left(x^2+2x.\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{3}\right)\)
\(=-3\left(x+\dfrac{1}{3}\right)^2-1\le1\)
Dấu " = " khi \(-3\left(x+\dfrac{1}{3}\right)^2=0\Leftrightarrow x=\dfrac{-1}{3}\)
Vậy \(MAX_Q=1\) khi \(x=\dfrac{-1}{3}\)
