Rút gọn biểu thức: \(P = \left( {\dfrac{{3\sqrt x }}{{\sqrt x + 2}} + \dfrac{{\sqrt x }}{{\sqrt x - 2}} - \dfrac{{x - \sqrt x }}{{x - 4}}} \right):\dfrac{{3\sqrt x }}{{\sqrt x +...

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Điều kiện: \(x > 0,\;\;x \ne 4.\)

\(\begin{array}{l}P = \left( {\dfrac{{3\sqrt x }}{{\sqrt x  + 2}} + \dfrac{{\sqrt x }}{{\sqrt x  - 2}} - \dfrac{{x - \sqrt x }}{{x - 4}}} \right):\dfrac{{3\sqrt x }}{{\sqrt x  + 2}}\\\;\;\; = \left( {\dfrac{{3\sqrt x }}{{\sqrt x  + 2}} + \dfrac{{\sqrt x }}{{\sqrt x  - 2}} - \dfrac{{x - \sqrt x }}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}} \right):\dfrac{{3\sqrt x }}{{\sqrt x  + 2}}\\\;\;\; = \dfrac{{3\sqrt x \left( {\sqrt x  - 2} \right) + \sqrt x \left( {\sqrt x  + 2} \right) - x + \sqrt x }}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}:\dfrac{{3\sqrt x }}{{\sqrt x  + 2}}\\\;\;\; = \dfrac{{3x - 6\sqrt x  + x + 2\sqrt x  - x + \sqrt x }}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}.\dfrac{{\sqrt x  + 2}}{{3\sqrt x }}\\\;\;\; = \dfrac{{3x - 3\sqrt x }}{{\sqrt x  - 2}}.\dfrac{1}{{3\sqrt x }} = \dfrac{{3\sqrt x \left( {\sqrt x  - 1} \right)}}{{3\sqrt x \left( {\sqrt x  - 2} \right)}} = \dfrac{{\sqrt x  - 1}}{{\sqrt x  - 2}}.\end{array}\)