\(\dfrac{-5}{x^2-2x+3}\)=\(\dfrac{-5}{\left(x^2-2x+1\right)+2}\)=\(\dfrac{-5}{\left(x-1\right)^2+2}\)
Vì (x-1)² +2 ≥ 2 ∀ x ∈ R
⇒ \(\dfrac{-5}{\left(x-1\right)^2+2}\ge\dfrac{-5}{2}\)
⇒ GTNN của biểu thức ≥ \(\dfrac{-5}{2}\)

⇒ GTNN = \(\dfrac{-5}{2}\)
Khi x-1= 0 ⇒ x=1