\(A=\left(x^2+3x+4\right)^2\)
ta có:
\(x^2+3x+4=x^2+2\cdot\dfrac{3}{2}x+\left(\dfrac{3}{2}\right)^2+\dfrac{7}{4}\\ =\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
vậy \(minA=\left(\dfrac{7}{4}\right)^2=\dfrac{49}{16}\Leftrightarrow x=-\dfrac{3}{2}\)