Ta có: \(P=\frac{2x+3}{\sqrt{x}-2}\)
\(=\frac{2x-8+11}{\sqrt{x}-2}\)
\(=\frac{2\left(x-4\right)+11}{\sqrt{x}-2}=2\left(\sqrt{x}+2\right)+\frac{11}{\sqrt{x}-2}\)
\(=2\left(\sqrt{x}-2+4\right)+\frac{11}{\sqrt{x}-2}=2\left(\sqrt{x}-2\right)+\frac{11}{\sqrt{x}-2}+8\)
=>\(P\ge2\cdot\sqrt{2\left(\sqrt{x}-2\right)\cdot\frac{11}{\sqrt{x}-2}}+8=2\sqrt{22}+8\)
Dấu '=' xảy ra khi \(2\left(\sqrt{x}-2\right)=\frac{11}{\left(\sqrt{x}-2\right)}\)
=>\(\left(\sqrt{x}-2\right)^2=\frac{11}{2}=\frac{22}{4}\)
=>\(\sqrt{x}-2=\frac{\sqrt{22}}{2}\)
=>\(\sqrt{x}=\frac{\sqrt{22}+4}{2}\)
=>\(x=\frac{16+22+8\sqrt{22}}{4}=\frac{38+8\sqrt{22}}{4}=\frac{19+4\sqrt{22}}{2}\)
