Cho biểu thức \(P = \left( {\dfrac{{2x + 1}}{{\sqrt {{x^3}}  - 1}} - \dfrac{1}{{\sqrt x  - 1}}} \right):\left( {1 - \dfrac{{x + 4}}{{x + \sqrt x  + 1}}} \right)\)

Rút gọn P.

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\x \ne 1\end{array} \right.\)

\(\begin{array}{l}P = \left( {\dfrac{{2x + 1}}{{\sqrt {{x^3}}  - 1}} - \dfrac{1}{{\sqrt x  - 1}}} \right):\left( {1 - \dfrac{{x + 4}}{{x + \sqrt x  + 1}}} \right)\\ = \left( {\dfrac{{2x + 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} - \dfrac{1}{{\sqrt x  - 1}}} \right):\left( {\dfrac{{x + \sqrt x  + 1 - x - 4}}{{x + \sqrt x  + 1}}} \right)\\ = \dfrac{{2x + 1 - x - \sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}:\dfrac{{\sqrt x  - 3}}{{x + \sqrt x  + 1}}\\ = \dfrac{{x - \sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}.\dfrac{{x + \sqrt x  + 1}}{{\sqrt x  - 3}}\,\,\,\,\left( {x \ne 9} \right)\\ = \dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 3} \right)}} = \dfrac{{\sqrt x }}{{\sqrt x  - 3}}.\end{array}\)    

Vậy \(P = \dfrac{{\sqrt x }}{{\sqrt x  - 3}}\) với \(x \ge 0;x \ne 1;x \ne 9\)