Ta có : \(x\left(x+1\right)+\dfrac{3}{2}=x^2+x+\dfrac{3}{2}=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{5}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\ge\dfrac{5}{4}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy Min của b/t trên là : \(\dfrac{5}{4}\Leftrightarrow x=-\dfrac{1}{2}\)