\(A=\frac{x^2-1}{x^2+1}=\frac{x^2+1-2}{x^2+1}=1-\frac{2}{x^2+1}\)
Ta có:
\(x^2+1\ge1\forall x\Rightarrow\frac{2}{x^2+1}\le\frac{2}{1}\forall x\Rightarrow-\frac{2}{x^2+1}\ge-2\)
\(A=1-\frac{2}{x^2+1}\ge1-2=-1\)
Dấu " = " xảy ra <=> x=0
Vậy \(A_{min}=-1\Leftrightarrow x=0\)