a/ đk: x>0; \(x\ne1\)
A= \(\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
=\(\left[\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\) : \(\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\left[\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right]:\)\(\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
=\(\frac{2\sqrt{x}}{\sqrt{x}}:\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=2.\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
vậy A =\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) với x>0; \(x\ne1\)
b/để A<0 thì: \(\frac{\sqrt{x}+1}{\sqrt{x}-1}< 0\)
mà x>0; \(x\ne1\)=> \(\sqrt{x}+1>0\)
nên để A<0 thì \(\sqrt{x}-1< 0\)=> x<1 kết hợp với đkxđ
=> \(0< x< 1\)
vậy \(0< x< 1\) thì A<0
