Đk: `x >=0`.
`sqrt x >=0 -> sqrt x + 3 >= 0 + 3 forall x in RR`
`-> 3/(sqrtx + 3) >=3/3 = 1 forall x in RR`
Dấu bằng xảy ra `<=> x = 0`.
\(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\dfrac{3}{\sqrt{x}+3}\le\dfrac{3}{3}=1\)\(\Rightarrow\) Max =1\(\Leftrightarrow x=0\)