\(=\left(9x^2-6x+1\right)+4=\left(3x-1\right)^2+4\ge4\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{3}\)
\(9x^2-6x+5\\ =9\left(x^2-\dfrac{2}{3}x+\dfrac{5}{9}\right)\)
\(=9\).(\(x^2-2.x\).\(\dfrac{1}{3}\)\(+\dfrac{1}{9}+\dfrac{4}{9}\))
\(=9\left(x-\dfrac{1}{3}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\)
Dấu "=" xảy ra khi \(9\left(x-\dfrac{1}{3}\right)^2=0\Leftrightarrow\left(x-\dfrac{1}{3}\right)^2=0\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)\(\)