a,Thay x\(=\dfrac{1}{4}\) vào B ta có:
\(B=\dfrac{x-\sqrt{x}}{\sqrt{x}-2}=\dfrac{\dfrac{1}{4}-\sqrt{\dfrac{1}{4}}}{\sqrt{\dfrac{1}{4}}-2}=\dfrac{\dfrac{1}{4}-\dfrac{1}{2}}{\dfrac{1}{2}-2}=\dfrac{-\dfrac{1}{4}}{-\dfrac{3}{2}}=\dfrac{1}{6}\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-x}\right).\dfrac{x-2\sqrt{x}+1}{\sqrt{x}+1}\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(1-\sqrt{x}\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
