ĐK: $x\ge 0$
$P\,=\dfrac{x+7}{\sqrt x+3}\\\quad =\dfrac{x+3\sqrt x-3\sqrt x-9+16}{\sqrt x+3}\\\quad =\dfrac{(x+3\sqrt x)-(3\sqrt x+9)+16}{\sqrt x+3}\\\quad =\dfrac{\sqrt x(\sqrt x+3)-3(\sqrt x+3)+16}{\sqrt x+3}\\\quad =\sqrt x-3+\dfrac{16}{\sqrt x+3}\\\quad =(\sqrt x+3)+\dfrac{16}{\sqrt x+3}-6$
$\sqrt x\ge 0\\↔\sqrt x+3\ge 0\\↔\sqrt x+3>0\\→\dfrac{16}{\sqrt x+3}>0$
Áp dụng bất đẳng thức Cô-si cho 2 số dương $\sqrt x+3$ và $\dfrac{16}{\sqrt x+3}$
$(\sqrt x+3)+\dfrac{16}{\sqrt x+3}\ge 2\sqrt{(\sqrt x+3).\dfrac{16}{\sqrt x+3}}\\↔(\sqrt x+3)+\dfrac{16}{\sqrt x+3}\ge 2\sqrt{16}\\↔(\sqrt x+3)+\dfrac{16}{\sqrt x+3}\ge 8\\↔P\ge 2$
$→$ Dấu "=" xảy ra khi $\sqrt x+3=\dfrac{16}{\sqrt x+3}$
$↔(\sqrt x+3)^2=16\\↔\sqrt x+3=4(vì\,\,\sqrt x+3>0)\\↔\sqrt x=1\\↔x=1(TM)$
Vậy $P$ đạt GTNN là $2$ tại $x=1$
