Bài 2:
a: ĐKXĐ: x>0; x<>1
\(A=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\cdot\left(\dfrac{1-\sqrt{x}}{2\sqrt{x}}\right)^2\)
\(=\dfrac{-4\sqrt{x}}{x-1}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{4x}\)
\(=\dfrac{-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
b: Để \(\dfrac{A}{\sqrt{x}}>2\) thì \(\dfrac{-\sqrt{x}+1}{x\left(\sqrt{x}+1\right)}-2>0\)
\(\Leftrightarrow-\sqrt{x}+1-2x\left(\sqrt{x}+1\right)>0\)
\(\Leftrightarrow-2x\sqrt{x}-2x-\sqrt{x}+1>0\)
Đến đây thì xin lỗi bạn, mình bí rồi