Đáp án:
$S=∅$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} ĐKXĐ:\ x\neq 1;\ x >0\\ P=\left[\frac{\left(\sqrt{x} -1\right)\left( x+\sqrt{x} +1\right)}{\sqrt{x}\left(\sqrt{x} -1\right)} -\frac{\left(\sqrt{x} +1\right)\left( x-\sqrt{x} +1\right)}{\sqrt{x}\left(\sqrt{x} +1\right)}\right] .\frac{x-1}{2\left(\sqrt{x} -1\right)^{2}}\\ P=\left(\frac{x+\sqrt{x} +1}{\sqrt{x}} -\frac{x-\sqrt{x} +1}{\sqrt{x}}\right) .\frac{\sqrt{x} +1}{2}\\ P=\frac{x+\sqrt{x} +1-x+\sqrt{x} -1}{\sqrt{x}} .\frac{\sqrt{x} +1}{2}\\ P=\frac{2\sqrt{x}}{\sqrt{x}}\frac{\sqrt{x} +1}{2} =\sqrt{x} +1\\ Ta\ có\ P< 0\Leftrightarrow \sqrt{x} +1< 0\Leftrightarrow \sqrt{x} < \ -1\ ( vô\ lý)\\ Vậy\ S=\emptyset \end{array}$
