Ta có:
$\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\geq 10-6=4$
Dấu '=' xảy ra khi $x=4$
Vậy...
Ta có:
$\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\geq 10-6=4$
Dấu '=' xảy ra khi $x=4$
Vậy...