a: \(A=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
b: Khi x=4+2căn 3 thì \(A=\dfrac{1}{\sqrt{3}+1+2}=\dfrac{1}{3+\sqrt{3}}=\dfrac{3-\sqrt{3}}{6}\)
c: A>1/6
=>A-1/6>0
=>\(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{6}>0\)
=>\(6-\sqrt{x}-2>0\)
=>4-căn x>0
=>0<x<16

