Rút gọn biểu thức: $A = \left( {\dfrac{{\sqrt x }}{2} - \dfrac{1}{{2\sqrt x }}} \right)\left( {\dfrac{{x - \sqrt x }}{{\sqrt x  + 1}} - \dfrac{{x + \sqrt x }}{{\sqrt x  - 1}}} \right)$ v ới $x > 0;\,\,x \ne 1.$

$\begin{array}{l}A = \left( {\dfrac{{\sqrt x }}{2} - \dfrac{1}{{2\sqrt x }}} \right)\left( {\dfrac{{x - \sqrt x }}{{\sqrt x  + 1}} - \dfrac{{x + \sqrt x }}{{\sqrt x  - 1}}} \right) = \dfrac{{x - 1}}{{2\sqrt x }}.\left( {\dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\sqrt x  - 1}}} \right)\\\,\,\,\,\,\, = \dfrac{{x - 1}}{{2\sqrt x }}.\dfrac{{\sqrt x {{\left( {\sqrt x  - 1} \right)}^2} - \sqrt x {{\left( {\sqrt x  + 1} \right)}^2}}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}} = \dfrac{{x - 1}}{{2\sqrt x }}.\dfrac{{\sqrt x \left[ {x - 2\sqrt x  + 1 - \left( {x + 2\sqrt x  + 1} \right)} \right]}}{{x - 1}}\\\,\,\,\,\,\, = \dfrac{{x - 2\sqrt x  + 1 - x - 2\sqrt x  - 1}}{2} = \dfrac{{ - 4\sqrt x }}{2} =  - 2\sqrt x .\end{array}$

Vậy $A =  - 2\sqrt x$ với $x > 0;\,\,x \ne 1.$