Điều kiện xác định $x\ge 0, x\ne 1$
$\begin{array}{l} \left( {\dfrac{{\sqrt x - 2}}{{x - 1}} - \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\dfrac{{{{\left( {1 - x} \right)}^2}}}{2}\\ = \left[ {\dfrac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right].\dfrac{{{{\left( {1 - x} \right)}^2}}}{2}\\ = \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}\left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\ = \dfrac{{x - \sqrt x - 2 - \left( {x + \sqrt x - 2} \right)}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\ = \dfrac{{ - 2\sqrt x }}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\ = \dfrac{{ - \sqrt x \left( {x - 1} \right)}}{{\sqrt x + 1}} = \dfrac{{ - \sqrt x \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\ = - \sqrt x \left( {\sqrt x - 1} \right) = \sqrt x - x \end{array}$
