\(R=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
\(\Rightarrow R\ge2\sqrt{\left(\sqrt{x}-1\right).\frac{2}{\left(\sqrt{x}-1\right)}}+3=3+2\sqrt{2}\)
\(\Rightarrow R_{min}=3+2\sqrt{2}\) khi \(\sqrt{x}-1=\sqrt{2}\Rightarrow x=3+2\sqrt{2}\)