\(=5\left(x^2-\dfrac{12}{5}x+\dfrac{1}{5}\right)\)
\(=5\left(x^2-2\cdot x\cdot\dfrac{6}{5}+\dfrac{36}{25}-\dfrac{31}{25}\right)\)
\(=5\left(x-\dfrac{6}{5}\right)^2-\dfrac{31}{5}>=-\dfrac{31}{5}\)
Dấu '=' xảy ra khi x=6/5
\(f\left(x\right)=5\left(x^2-\dfrac{2.6}{5}x+\dfrac{36}{25}-\dfrac{36}{25}\right)+1=5\left(x-\dfrac{6}{5}\right)^2-\dfrac{31}{5}\ge-\dfrac{31}{5}\)
Dấu ''='' xảy ra khi x = 6/5
Ta có:
\(f\left(x\right)=5x^2-12x+1=5\left(x^2-\dfrac{12}{5}x+\dfrac{1}{5}\right)=5\left[\left(x^2-2.\dfrac{6}{5}x+\dfrac{36}{25}\right)-\dfrac{36}{25}+\dfrac{1}{5}\right]=5\left[\left(x-\dfrac{6}{5}\right)^2-\dfrac{31}{25}\right]=5\left(x-\dfrac{6}{5}\right)^2-\dfrac{31}{5}\ge-\dfrac{31}{5}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{6}{5}\)
Vậy GTNN của f(x) là \(-\dfrac{31}{5}\) khi \(x=\dfrac{6}{5}\)
