Tìm \(x\) để \(P =  - 1.\)

Điều kiện: \(x > 0,x \ne 4,x \ne 9\)

\(\begin{array}{l}P = \left( {\dfrac{{4\sqrt x }}{{2 + \sqrt x }} + \dfrac{{8x}}{{4 - x}}} \right):\left( {\dfrac{{\sqrt x  - 1}}{{x - 2\sqrt x }} - \dfrac{2}{{\sqrt x }}} \right)\\ = \left( {\dfrac{{4\sqrt x }}{{2 + \sqrt x }} + \dfrac{{8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}} \right):\left( {\dfrac{{\sqrt x  - 1}}{{\sqrt x \left( {\sqrt x  - 2} \right)}} - \dfrac{2}{{\sqrt x }}} \right)\\ = \dfrac{{4\sqrt x \left( {2 - \sqrt x } \right) + 8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}:\dfrac{{\sqrt x  - 1 - 2\left( {\sqrt x  - 2} \right)}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}\\ = \dfrac{{8\sqrt x  + 4x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  - 2} \right)}}{{3 - \sqrt x }}\\ = \dfrac{{4\sqrt x \left( {2 + \sqrt x } \right)}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {2 - \sqrt x } \right)}}{{\sqrt x  - 3}}\\ = \dfrac{{4x}}{{\sqrt x  - 3}}\end{array}\)

Với điều kiện: \(x > 0,x \ne 4,x \ne 9\).

Ta có: \(P =  - 1\)

\(\begin{array}{l} \Leftrightarrow \dfrac{{4x}}{{\sqrt x  - 3}} =  - 1 \Leftrightarrow 4x + \sqrt x  - 3 = 0 \Leftrightarrow 4x + 4\sqrt x  - 3\sqrt x  - 3 = 0 \Leftrightarrow 4\sqrt x \left( {\sqrt x  + 1} \right) - 3\left( {\sqrt x  + 1} \right) = 0\\ \Leftrightarrow \left( {\sqrt x  + 1} \right)\left( {4\sqrt x  - 3} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}\sqrt x  =  - 1\left( {ktm} \right)\\\sqrt x  = \dfrac{3}{4} \Leftrightarrow x = \dfrac{9}{{16}}\left( {tm} \right)\end{array} \right.\end{array}\)

Vậy \(x = \dfrac{9}{{16}}\)  thì \(P =  - 1.\)