Question

Answer 1

\(-\left(x^2+2x+4\right)\le-3\)

\(\Leftrightarrow A\ge-\dfrac{1}{3}\forall x\)

Dấu '=' xảy ra khi x=-1

Answer 2

\(A=\dfrac{1}{-\left(x^2-2x+1\right)-3}=\dfrac{1}{-\left(x-1\right)^2-3}\)

Ta có \(-\left(x-1\right)^2-3\le-3\Leftrightarrow A\ge\dfrac{1}{-3}=-\dfrac{1}{3}\)

Dấu \("="\Leftrightarrow x=1\)