Ta có : \(B=\frac{14x^2-8x+9}{3x^2+6x+9}=\frac{2\left(x^2+2x+3\right)+\left(12x^2-12x+3\right)}{3\left(x^2+2x+3\right)}\)
\(=\frac{12\left(x-\frac{1}{2}\right)^2}{3\left(x^2+2x+3\right)}+\frac{2}{3}\ge\frac{2}{3}\) . Dấu "=" xảy ra khi x = 1/2
Vậy Min B = 2/3 khi x = 1/2