a) \(A=x^2-2x-1\)
\(=x^2-2.x.1+1^2-1-1\)
\(=\left(x-1\right)^2-2\)
Vì \(\left(x-1\right)^2\ge0\forall x\)
nên \(\left(x-1\right)^2-2\ge-2\forall x\)
\(\Rightarrow A\ge-2\forall x\)
Dấu "=" xảy ra khi \(\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy \(MIN_A=-2\) khi \(x=1.\)
b) \(B=x^2+3x+7\)
\(=x^2+2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2-\dfrac{9}{4}+7\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\)
\(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\forall x\)
\(\Rightarrow B\ge\dfrac{19}{4}\forall x\)
Dấu "=" xảy ra khi \(\left(x+\dfrac{3}{2}\right)^2=0\Rightarrow x=-\dfrac{3}{2}\)
Vậy \(MIN_B=\dfrac{19}{4}\) khi \(x=-\dfrac{3}{2}.\)
