Sửa đề: \(A=\dfrac{5}{2\sqrt{x}+1}+\dfrac{3}{2\sqrt{x}-1}+\dfrac{12\sqrt{x}}{1-4x}\)
a: \(A=\dfrac{10\sqrt{x}-5+6\sqrt{x}+3-12\sqrt{x}}{4x-1}\)
\(=\dfrac{4\sqrt{x}-2}{4x-1}=\dfrac{2\left(2\sqrt{x}-1\right)}{4x-1}=\dfrac{2}{2\sqrt{x}+1}\)
b: Để A>1/3 thì A-1/3>0
\(\Leftrightarrow\dfrac{2}{2\sqrt{x}+1}-\dfrac{1}{3}>0\)
\(\Leftrightarrow6-2\sqrt{x}-1>0\)
\(\Leftrightarrow5-2\sqrt{x}>0\)
=>\(2\sqrt{x}< 5\)
=>x<25/4
Vậy: 0<x<25/4 và x<>1/4