1. Thay x = 64 (tmđk) vào A :
\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{5}{4}\)
Vậy : Khi x = 64, \(A=\dfrac{5}{4}\).
2. \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
Vậy : \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\).
3. Ta có : \(\dfrac{A}{B}=\dfrac{\dfrac{2+\sqrt{x}}{\sqrt{x}}}{\dfrac{\sqrt{x}+2}{\sqrt{x}+1}}\)
\(=\dfrac{2+\sqrt{x}}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\dfrac{2+\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{\left(2+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Để \(\dfrac{A}{B}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
\(\Leftrightarrow\dfrac{2\left(\sqrt{x}+1\right)-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Leftrightarrow2\left(\sqrt{x}+1\right)-3\sqrt{x}>0\)
\(\Leftrightarrow2\sqrt{x}+2-3\sqrt{x}>0\)
\(\Leftrightarrow-\sqrt{x}>-2\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ : Vậy : \(\dfrac{A}{B}>\dfrac{3}{2}\) khi và chỉ khi \(0< x< 4\)
