\(f\left(x\right)=\dfrac{x-1}{2}+\dfrac{2}{x-1}+\dfrac{1}{2}\ge2\sqrt{\dfrac{\left(x-1\right)}{2}.\dfrac{2}{\left(x-1\right)}}+\dfrac{1}{2}=\dfrac{5}{2}\)
\(\Rightarrow f\left(x\right)_{min}=\dfrac{5}{2}\) khi \(\dfrac{x-1}{2}=\dfrac{2}{x-1}\Rightarrow x=3\)