Ta có : \(Q=2x-2-3x^2=-\left(3x^2-2x+2\right)=-[3\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+\frac{17}{9}]\)
\(=-[3\left(x-\frac{1}{3}\right)^2+\frac{17}{9}]\)
Ta có : \(\left(x-\frac{1}{3}\right)^2\ge0=>-[3\left(x-\frac{1}{3}\right)^2+\frac{17}{9}]\ge0\)
Dấu bằng xảy ra khi \(x-\frac{1}{3}=0=>x=\frac{1}{3}\)
Vậy \(Q_{max}=\frac{17}{9}\)khi \(x=\frac{1}{3}\)