\(\dfrac{A}{B}=\dfrac{x^3-1}{x-1}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}=x^2+x+1\)
Ta có :
\(x^2+x+1\)
\(=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN của \(\dfrac{A}{B}\) là : \(\dfrac{3}{4}\)
Dấu = xảy ra khi : \(x=-\dfrac{1}{2}\)