a . ĐKXĐ : \(x>0;x\ne1\)
Ta có : \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\) . \(\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\left[\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right].\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
\(=2.\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) . Vậy ,,,
b. \(A< 0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow0< x< 1\)
c. \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\) A \(\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\in Z\)
Mà \(\sqrt{x}-1>-1\) . Do đó : \(\sqrt{x}-1\in\left\{1;2\right\}\Leftrightarrow x\in\left\{4;9\right\}\)
Vậy ...
