Tìm Max
Đặt A=\(2x-2-3x^2\)
\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{5}{3}\)
\(=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{5}{3}\)
Với mọi giá trị của x ta có:
\(\left(x-\dfrac{1}{3}\right)^2\ge0\Rightarrow-3\left(x-\dfrac{1}{3}\right)^2\le0\)
\(\Rightarrow-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{5}{3}\le-\dfrac{5}{3}\)
Vậy Max A = \(-\dfrac{5}{3}\) khi \(x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)