a/ \(A=x^2+3x+7=\left(x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}\right)+\dfrac{19}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\)
Vì: \(\left(x+\dfrac{3}{2}\right)^2\ge0\Rightarrow\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
dấu ''='' xảy ra khi x = \(-\dfrac{3}{2}\)
vậy MinA = \(\dfrac{19}{4}\Leftrightarrow x=-\dfrac{3}{2}\)
b/ \(B=x^2+5x+7=\left(x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)
vì: \(\left(x+\dfrac{5}{2}\right)^2\ge0\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
dấu ''='' xảy ra khi x = \(-\dfrac{5}{2}\)
vậy MinB = \(\dfrac{3}{4}\)
c/ \(C=x\left(x-6\right)=x^2-6x=\left(x^2-6x+9\right)-9=\left(x-3\right)^2-9\)
vì: \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2-9\ge-9\)
dấu ''='' xảy ra khi \(x=3\)
vậy MinC = - 9
