a. A= \(x^2-4x+5=\left(x-2\right)^2+1\ge1\)
Vậy minA=1<=> x=2
b. B=\(2x^2-4x-6=\left(x\sqrt{2}-\sqrt{2}\right)^2-8\ge-8\)
Vậy minB=-8 <=> x=1
c. C=\(3x^2+9x+6=\left(\sqrt{3}x+\frac{\sqrt{3}}{2}\right)^2-\frac{3}{4}\ge\frac{-3}{4}\)
Vậy minC=-3/4 <=> x=-3/2
d. D=\(5x^2+5x+1=\left(\sqrt{5}x+\frac{\sqrt{5}}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\)
Vậy minD=-1/4 <=> x=-1/2