đk: x>0
để\(A\ge\frac{1}{3\sqrt{x}}thì\):\(\frac{1}{\sqrt{x\left(1+\sqrt{x}\right)}}-\frac{1}{3\sqrt{x}}\ge0\)
\(\Leftrightarrow\frac{3\sqrt{x}}{3\sqrt{x}.\sqrt{x\left(1+\sqrt{x}\right)}}-\frac{\sqrt{x\left(1+\sqrt{x}\right)}}{3\sqrt{x}.\sqrt{x\left(1+\sqrt{x}\right)}}\ge0\)
\(\Leftrightarrow3\sqrt{x}-\sqrt{x\left(1+\sqrt{x}\right)}\ge0\)
\(\Leftrightarrow3\sqrt{x}\ge.\sqrt{x\left(1+\sqrt{x}\right)}\)
\(\Leftrightarrow9x\ge x\left(1+\sqrt{x}\right)\)
\(\Leftrightarrow x\left(8-\sqrt{x}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\le64\end{matrix}\right.\)kết hợp với đkxđ
vậy x>0 hoặc 0<x\(\le64\) thỏa mãn đk đề bài
