\(A=4x^4+4x^2-3\)
\(A=\left[\left(2x^2\right)^2+2.2x^2.1+1^2\right]-4\)
\(A=\left(2x+1\right)^2-4\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+1\right)^2-4\ge-4\forall x\)
\(A=-4\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy \(A_{min}=-4\Leftrightarrow x=-\frac{1}{2}\)