ĐKXĐ: x>=0
\(\dfrac{x+5}{\sqrt{x}+2}=\dfrac{x-4+9}{\sqrt{x}+2}=\sqrt{x}-2+\dfrac{9}{\sqrt{x}+2}\)
\(=\sqrt{x}+2+\dfrac{9}{\sqrt{x}+2}-4>=2\cdot\sqrt{\left(\sqrt{x}+2\right)\cdot\dfrac{9}{\sqrt{x}+2}}-4=2\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x}+2=\sqrt{9}=3\)
=>\(\sqrt{x}=3-2=1\)
=>x=1


