Cho biểu thức \(P = \left( {\frac{{x - 6}}{{x + 3\sqrt x }} - \frac{1}{{\sqrt x }} + \frac{1}{{\sqrt x + 3}}} \right):\frac{{2\sqrt x - 6}}{{x + 1}}\) với \(x > 0,\;\;x \ne 9.\)...

* Hướng dẫn giải

Điều kiện: \(x > 0,\;x \ne 9.\)

\(\begin{array}{l}P = \left( {\dfrac{{x - 6}}{{x + 3\sqrt x }} - \dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt x  + 3}}} \right):\dfrac{{2\sqrt x  - 6}}{{x + 1}}\\\;\;\; = \left( {\dfrac{{x - 6}}{{\sqrt x \left( {\sqrt x  + 3} \right)}} - \dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt x  + 3}}} \right):\dfrac{{2\left( {\sqrt x  - 3} \right)}}{{x + 1}}\\\;\;\; = \dfrac{{x - 6 - \left( {\sqrt x  + 3} \right) + \sqrt x }}{{\sqrt x \left( {\sqrt x  + 3} \right)}}.\dfrac{{x + 1}}{{2\left( {\sqrt x  - 3} \right)}}\\\;\;\; = \dfrac{{x - 6 - \sqrt x  - 3 + \sqrt x }}{{\sqrt x \left( {\sqrt x  + 3} \right)}}.\dfrac{{x + 1}}{{2\left( {\sqrt x  - 3} \right)}}\\\;\;\; = \dfrac{{\left( {x - 9} \right)\left( {x + 1} \right)}}{{2\sqrt x \left( {x - 9} \right)}} = \dfrac{{x + 1}}{{2\sqrt x }}.\end{array}\)