Đặt A = 4x2 + 12x = 4x2 + 12x + 9 - 9
= (2x + 3)2 - 9 \(\ge\)-9
Dấu " = " xảy ra <=> 2x + 3 = 0
<=> x = -1,5
Vậy Min A = -9 <=> x = -1,5
\(A=4x^2+12x=4x^2+12x+9-9=\left(2x+3\right)^2-9\ge-9\)
Dấu \(=\)khi \(2x+3=0\Leftrightarrow x=-\frac{3}{2}\).
Vậy \(minA=-9\).