\(A=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\\ A_{min}=3\Leftrightarrow x=-2\\ B=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\\ B_{min}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\\ D=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\\ D_{max}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)
a: \(A=x^2+4x+7=\left(x+2\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi x=-2
b: \(B=x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '='xảy ra khi \(x=\dfrac{1}{2}\)
