với x\(\ge\)0, ta có:
\(A=\frac{x+\sqrt{x}+4}{\sqrt{x}}=\frac{x}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}}+\frac{4}{\sqrt{x}}=\sqrt{x}+1+\frac{4}{\sqrt{x}}=1+\sqrt{x}+\frac{4}{\sqrt{x}}\)
mà \(\sqrt{x}+\frac{4}{\sqrt{x}}\ge2\sqrt{\sqrt{x}\times\frac{4}{\sqrt{x}}}=2\sqrt{4}=4\)
\(\Rightarrow1+\sqrt{x}+\frac{4}{\sqrt{x}}\ge1+4=5\)
Dấu = xảy ra khi \(\sqrt{x}=\frac{4}{\sqrt{x}}\Leftrightarrow x=4\left(tm\right)\)
vậy MinA=5 khi x=4