a, B=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x-1}\)=\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\sqrt{x}}{x-1}\)
B = \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}}{x-1}=\dfrac{x+\sqrt{x}-2\sqrt{x}}{x-1}\)
B=\(\dfrac{x-\sqrt{x}}{x-1}=\dfrac{\sqrt{x}\left(x-1\right)}{x-1}=\sqrt{x}\)
C = \(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}=\dfrac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
C=\(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\) \(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có : \(B.C=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}.\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}-1=\dfrac{1}{3}\Leftrightarrow\sqrt{x}=\dfrac{1}{3}+1=\dfrac{4}{3}\)
\(\Leftrightarrow x=\dfrac{16}{9}\)
