a: \(A=x^2-4x+12\)
\(=x^2-4x+4+8\)
\(=\left(x-2\right)^2+8\ge8\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2
b: \(B=4x^2-10x-15\)
\(=4x^2-2\cdot2x\cdot\frac52+\frac{25}{4}-\frac{85}{4}\)
\(=\left(2x-\frac52\right)^2-\frac{85}{4}\ge-\frac{85}{4}\forall x\)
Dấu '=' xảy ra khi \(2x-\frac52=0\)
=>\(2x=\frac52\)
=>\(x=\frac54\)
c: \(C=x^2+x+4\)
\(=x^2+x+\frac14+\frac{15}{4}\)
\(=\left(x+\frac12\right)^2+\frac{15}{4}\ge\frac{15}{4}\forall x\)
Dấu '=' xảy ra khi \(x+\frac12=0\)
=>\(x=-\frac12\)
d: \(D=-x^2+6x-8\)
\(=-\left(x^2-6x+8\right)\)
\(=-\left(x^2-6x+9-1\right)=-\left(x-3\right)^2+1\le1\forall x\)
Dấu '=' xảy ra khi x-3=0
=>x=3
e: \(E=-4x^2+6x-7\)
\(=-\left(4x^2-6x+7\right)\)
\(=-\left(4x^2-2\cdot2x\cdot\frac32+\frac94+\frac{19}{4}\right)=-\left(2x-\frac32\right)^2-\frac{19}{4}\le-\frac{19}{4}\forall x\)
Dấu '=' xảy ra khi \(2x-\frac32=0\)
=>\(2x=\frac32\)
=>\(x=\frac34\)
f: \(F=-2x^2-5x+2\)
\(=-2\left(x^2+\frac52x-1\right)\)
\(=-2\left(x^2+2\cdot x\cdot\frac54+\frac{25}{16}-\frac{41}{16}\right)\)
\(=-2\left(x+\frac54\right)^2+\frac{41}{8}\le\frac{41}{8}\forall x\)
Dấu '=' xảy ra khi \(x+\frac54=0\)
=>\(x=-\frac54\)
