\(B=-3x^2-12x-8=-3\left(x^2+4x+4\right)+4=-3\left(x+2\right)^2+4\le4\)
Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).
B=-3x^2-12x-8
Ta có:-3x^2-12x-8
=-(3x^2+2.3x.2+4)+4
=-(3x+2)^2+4
Vì : (3x+2)^2 > 0
=> -(3x+2)^2 < 0
=>-(3x+2)^2+4< 4
Dấu '=' xảy ra khi (3x+2)^2=0
=>3x+2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
Vậy Bmin=4 khi x=-2/3