Question

\(a,2x-2-3x^2\)

Answer 1

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 \(x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)