\(f\left(x\right)=3\left(x^2-\dfrac{2.4}{3}+\dfrac{16}{9}-\dfrac{16}{9}\right)=3\left(x-\dfrac{4}{3}\right)^2-\dfrac{16}{3}\ge-\dfrac{16}{3}\)
Dấu ''='' xảy ra khi x = 4/3
\(=3\left(x^2-\dfrac{8}{3}x+\dfrac{16}{9}-\dfrac{16}{9}\right)\)
\(=3\left(x-\dfrac{4}{3}\right)^2-\dfrac{16}{3}>=-\dfrac{16}{3}\)
Dấu '=' xảy ra khi x=4/3