C= \(9x^2-2x+3=\left(3x\right)^2-2.3x.\frac{1}{3}+\left(\frac{1}{3}\right)^2+3-\left(\frac{1}{3}\right)^2=\left(3x-\frac{1}{3}\right)^2+\frac{26}{9}\)
ta thấy: \(\left(3x-\frac{1}{3}\right)^2\ge0\)
=> C\(\ge\frac{26}{9}\)
=|> GTNN của C là \(\frac{26}{9}\)khi x=1/9