a: Thay x=64 vào A, ta được:
\(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)
b: Ta có: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
c: Ta có: \(\dfrac{A}{B}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Để \(\dfrac{A}{B}>\dfrac{3}{2}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
\(\Leftrightarrow2\sqrt{x}+2-3\sqrt{x}>0\)
\(\Leftrightarrow-\sqrt{x}>-2\)
hay x<4
Kết hợp ĐKXĐ, ta được: 0<x<4
