`P=((\sqrt{x}-2)/(x-1)-(\sqrt{x}+2)/(x+2\sqrt{x}+1))((1-x)/(\sqrt{2}))^2` Rút gọn ạ :'>

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