1: Thay x=64 vào A, ta được:
\(A=\dfrac{2+8}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)
2: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\cdot\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\cdot\sqrt{x}}=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
3: \(\dfrac{A}{B}>\dfrac{3}{2}\)
=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
=>\(\dfrac{2\left(\sqrt{x}+1\right)-3\sqrt{x}}{2\sqrt{x}}>0\)
=>\(2\sqrt{x}+2-3\sqrt{x}>0\)
=>\(-\sqrt{x}>-2\)
=>\(\sqrt{x}< 2\)
=>0<x<4