a: \(=2\left(x^2-4x+4-4\right)=2\left(x-2\right)^2-8>=-8\)
Dấu = xảy ra khi x=2
b: \(=5\left(x^2+4x+4-4\right)=5\left(x+2\right)^2-20>=-20\)
Dấu = xảy ra khi x=-2
c: \(=2\left(x^2-6x-\dfrac{7}{2}\right)\)
\(=2\left(x^2-6x+9-\dfrac{25}{2}\right)=2\left(x-3\right)^2-25>=-25\)
Dấu = xảy ra khi x=3
d: \(=3\left(x^2+6x+3\right)\)
\(=3\left(x^2+6x+9-6\right)\)
\(=3\left(x+3\right)^2-18>=-18\)
Dấu = xảy ra khi x=-3
c, B= \(2x^2-12x-7\)
B=2(\(x^2-6\)) -7
B= 2 ( \(x^2-2.x.2+4\))+2-7
B=2\(\left(x+2\right)^2-5\)
Vì \(\left(x+2\right)^2\)≥ 0 với mọi x ∈ R
<=>\(2\left(x+2\right)^2\) ≥0
<=>2\(\left(x+2\right)^2-5\) ≥ -5
<=> B≥-5
Vậy B\(_{min}\)=-5 khi x+2=0=>x=-2
