Điều kiện xác định: \(x \ne 1;x > 0\)

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

\( = 1:\dfrac{{x\sqrt x + x + 2\sqrt x + 2 + x\sqrt x + x - \sqrt x - 1 - \left( {x\sqrt x + x + \sqrt x + x + \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\)

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

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

+ So sánh \(P\) với \(3.\)

Xét \(P - 3 = \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} - 3 = \dfrac{{x + \sqrt x + 1 - 3\sqrt x }}{{\sqrt x }} = \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\)

Với \(x \ne 1;x > 0\) ta có: \(\sqrt x > 0\); $\sqrt x \ne 1$ nên \({\left( {\sqrt x - 1} \right)^2} > 0\) suy ra: $P-3 > 0$ hay $P > 3$