Rút gọn biểu thức $B$ ta được 

Ta có $B = \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( {x - 1} \right)}^2}}}{2}$

$= \left( {\dfrac{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right){{\left( {\sqrt x  + 1} \right)}^2}}} - \dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right){{\left( {\sqrt x  + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}{{\left( {\sqrt x  + 1} \right)}^2}}}{2}$

$= \dfrac{{x - \sqrt x  - 2 - x - \sqrt x  + 2}}{{\left( {\sqrt x  - 1} \right){{\left( {\sqrt x  + 1} \right)}^2}}}.\dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}.{{\left( {\sqrt x  + 1} \right)}^2}}}{2}$$= \dfrac{{ - 2\sqrt x \left( {\sqrt x  - 1} \right)}}{2} = \sqrt x  - x$

Vậy $B = \sqrt x  - x$.