\(A=\frac{14x^2-8x+9}{3x^2+6x+9}=\frac{2\left(x^2+2x+3\right)+3\left(4x^2+4x+1\right)}{3\left(x^2+2x+3\right)}=\frac{\left(2x+1\right)^2}{x^2+2x+3}+\frac{2}{3}\ge\frac{2}{3}\)
=> Min A = 2/3 khi x = -1/2
Mình lộn, phải là :
\(A=\frac{2\left(x^2+2x+3\right)+3\left(4x^2-4x+1\right)}{3\left(x^2+2x+3\right)}=\frac{\left(2x-1\right)^2}{\left(x^2+2x+3\right)}+\frac{2}{3}\ge\frac{2}{3}\)
\(\Rightarrow MinA=\frac{2}{3}\Leftrightarrow x=\frac{1}{2}\)