Ta có:
\(P=\frac{x+12}{\sqrt{x}+2}=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}\)
\(=\left(\sqrt{x}+2\right)+\frac{16}{\sqrt{x}+2}-4\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}-4=4\)
Dấu "=" xảy ra <=> \(\sqrt{x}+2=\frac{16}{\sqrt{x}+2}\Leftrightarrow\sqrt{x}+2=4\Leftrightarrow x=4\) thỏa mãn
=> min P = 4 tại x = 4.