A=[tex]\frac{x+4\sqrt{x}+5}{\sqrt{x}+2}[/tex]
ĐK: $x\ge 0$
$A=\dfrac{x+4\sqrt x+5}{\sqrt x+2}
\\=\sqrt x+2+\dfrac{1}{\sqrt x+2}
\\=\dfrac{\sqrt{x}+2}{4}+\dfrac{1}{\sqrt{x}+2}+\dfrac{3(\sqrt{x}+2)}{4}
\\\ge 2\sqrt{\dfrac{\sqrt x+2}{4}.\dfrac{1}{\sqrt x+2}}+\dfrac{3(0+2)}4
\\=1+\dfrac 32=\dfrac 52$
Dấu '=' xảy ra $\Leftrightarrow x=0$ (TM)
Vậy...