\(A=x^2+3x+7=x^2+2x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}+7\)
\(A=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu = xảy ra khi \(x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
vậy \(A_{min}=\dfrac{19}{4}\) khi \(x=-\dfrac{3}{2}\)
\(B=2\left(x^2+2x-\dfrac{1}{2}\right)=2\left(x^2+2x+1-1-\dfrac{1}{2}\right)\)
\(B=2\left[\left(x+1\right)^2-\dfrac{3}{2}\right]=2\left(x+1\right)^2-3\ge-3\)
Dấu = xảy ra khi x+1=0 <=> x=-1
vậy \(B_{min}=-3\) khi x=-1
a, \(A=x^2+3x+7\)
\(=\left[x^2+2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+7-\left(\dfrac{3}{2}\right)^2\)
\(=\left(x+\dfrac{3}{2}\right)^2+7-\dfrac{9}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\)
Vì \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\forall x\)
Dấu "=" xảy ra khi: \(\left(x+\dfrac{3}{2}\right)^2=0\)
\(\Leftrightarrow x+\dfrac{3}{2}=0\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy \(Min_A=\dfrac{19}{4}\Leftrightarrow x=-\dfrac{3}{2}\)
