Cho biểu thức \(P = \left( {\dfrac{{x - 2}}{{x + 2\sqrt x }} + \dfrac{1}{{\sqrt x  + 2}}} \right).\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\) với  \(x > 0;x \ne 1\)

 Tìm  \(x\)  để \(2P = 2\sqrt x  + 5\) .

Ta có: với \(x > 0,x \ne 1\)

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

Vậy với \(x > 0,x \ne 1\) ta có  \(P = \dfrac{{\sqrt x  + 1}}{{\sqrt x }}\)

Để \(2P = 2\sqrt x  + 5\)

Ta có: với \(x > 0,x \ne 1\)

\(\begin{array}{l}2P = 2\sqrt x  + 5\\ \Leftrightarrow 2.\dfrac{{\sqrt x  + 1}}{{\sqrt x }} = 2\sqrt x  + 5\\ \Rightarrow 2\sqrt x  + 2 = 2x + 5\sqrt x \\ \Leftrightarrow 2x + 3\sqrt x  - 2 = 0\\ \Leftrightarrow 2x - \sqrt x  + 4\sqrt x  - 2 = 0\\ \Leftrightarrow \sqrt x \left( {2\sqrt x  - 1} \right) + 2\left( {2\sqrt x  - 1} \right) = 0\\ \Leftrightarrow \left( {2\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt x  = \dfrac{1}{2} \Leftrightarrow x = \dfrac{1}{4}(tm)\\\sqrt x  =  - 2(ktm)\end{array} \right.\end{array}\)

Vậy \(x = \dfrac{1}{4}\) thì \(2P = 2\sqrt x  + 5\)