\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left[\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right]\)\(\left(x\ne1,x>0\right)\)
\(=\dfrac{ }{ }\)\(\left[\dfrac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)-\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}\right]\)\(:\left[\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(=\left[\dfrac{x^2\sqrt{x}+x^2-x-\sqrt{x}-x^2\sqrt{x}+x^2-x+\sqrt{x}}{x^2-x}\right]\)\(:\left[\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\right]\)
\(=\left[\dfrac{2x^2-2x}{x^2-x}\right].\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}=\dfrac{2.\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b, \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
A nguyên\(< =>1+\dfrac{2}{\sqrt{x}-1}\) nguyên\(< =>\sqrt{x}-1\in\) ước(2)\(=\left(\pm1;\pm2\right)\)
*\(\sqrt{x}-1=1=>x=4\left(TM\right)\)
*\(\sqrt{x}-1=-1< =>x=0\left(loai\right)\)
*\(\sqrt{x}-1=2=>x=9\left(TM\right)\)
*\(\sqrt{x}-1=-2< =>\sqrt{x}=-1\left(loai\right)\)
Vậy \(x\in\left\{4;9\right\}\) thì A nguyên