\(P=\dfrac{x}{\sqrt{x}-1}+2018=\dfrac{x-1+1}{\sqrt{x}-1}+2018\)
\(=\dfrac{x-1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}+2018=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}+2018\)
\(=\left(\sqrt{x}-1\right)+\dfrac{1}{\sqrt{x}-1}+2020\)
\(\ge2\sqrt{\left(\sqrt{x}-1\right).\dfrac{1}{\sqrt{x}-1}}+2020\) (BĐT Cauchy)
\(=2022\) (Dấu "=" khi \(\sqrt{x}-1=\dfrac{1}{\sqrt{x}-1}\Leftrightarrow x=4\) (tm))