a) \(A=2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\)
\(...B=\dfrac{6\sqrt{x}+1+\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}==\dfrac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}=\dfrac{2\sqrt{x}+1}{2\sqrt{x}-3}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}.\dfrac{2\sqrt{x}-3}{2\sqrt{x}+1}=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b) \(...\dfrac{A}{B}==\dfrac{3}{2}.\dfrac{\sqrt{x}-\dfrac{5}{3}}{\sqrt{x}+\dfrac{1}{2}}\)
Ta lại có \(\sqrt{x}\ge0\Rightarrow\sqrt{x}-\dfrac{5}{3}< \sqrt{x}+\dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}-\dfrac{5}{3}}{\sqrt{x}+\dfrac{1}{2}}< 1\)
\(\Rightarrow\dfrac{A}{B}< \dfrac{3}{2}\left(đpcm\right)\)
