Đáp án:

1) \(\dfrac{{\sqrt x  - 1}}{{\sqrt x }}\)

2) \(x = \dfrac{1}{4}\)

3) P=-1

4) \(3 - \sqrt 5 \)

5) \(0 < x < \dfrac{1}{9}\)

6) \(Min = \dfrac{3}{4}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)P = \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x  - 1} \right)}}:\dfrac{{\sqrt x  + 1}}{{{{\left( {\sqrt x  - 1} \right)}^2}}}\\
 = \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x  - 1} \right)}}.\dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}}}{{\sqrt x  + 1}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x }}\\
2)P =  - 1\\
 \to \dfrac{{\sqrt x  - 1}}{{\sqrt x }} =  - 1\\
 \to \sqrt x  - 1 =  - \sqrt x \\
 \to 2\sqrt x  = 1\\
 \to x = \dfrac{1}{4}\\
3)\left[ \begin{array}{l}
\sqrt x  - 1 = 0\\
2\sqrt x  - 1 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = \dfrac{1}{4}
\end{array} \right.\\
Thay:x = \dfrac{1}{4}\\
 \to P = \dfrac{{\sqrt {\dfrac{1}{4}}  - 1}}{{\sqrt {\dfrac{1}{4}} }} =  - 1\\
4)Thay:x = \sqrt {9 - 4\sqrt 5 }  + 3\sqrt 5  + 11\\
 = \sqrt {5 - 2.2.\sqrt 5  + 4}  + 3\sqrt 5  + 11\\
 = \sqrt {{{\left( {\sqrt 5  - 2} \right)}^2}}  + 3\sqrt 5  + 11\\
 = \sqrt 5  - 2 + 3\sqrt 5  + 11 = 4\sqrt 5  + 9\\
 = 5 + 2.2.\sqrt 5  + 4 = {\left( {\sqrt 5  + 2} \right)^2}\\
 \to P = \dfrac{{\sqrt {{{\left( {\sqrt 5  + 2} \right)}^2}}  - 1}}{{\sqrt {{{\left( {\sqrt 5  + 2} \right)}^2}} }}\\
 = \dfrac{{\sqrt 5  + 2 - 1}}{{\sqrt 5  + 2}} = 3 - \sqrt 5 \\
5)P < \dfrac{1}{{\sqrt x }} - 5\\
 \to \dfrac{{\sqrt x  - 1}}{{\sqrt x }} < \dfrac{1}{{\sqrt x }} - 5\\
 \to \dfrac{{\sqrt x  - 1 - 1 + 5\sqrt x }}{{\sqrt x }} < 0\\
 \to 6\sqrt x  - 2 < 0\left( {do:\sqrt x  > 0\forall x > 0} \right)\\
 \to x < \dfrac{1}{9}\\
 \to 0 < x < \dfrac{1}{9}\\
6)Q = \dfrac{1}{x} + 3P\\
 = \dfrac{1}{x} + 3.\dfrac{{\sqrt x  - 1}}{{\sqrt x }}\\
 = \dfrac{1}{x} + 3 - \dfrac{3}{{\sqrt x }}\\
Đặt:\dfrac{1}{{\sqrt x }} = t\left( {t > 0} \right)\\
 \to Q = {t^2} - 3t + 3\\
 = {t^2} - 2.t.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{3}{4}\\
 = {\left( {t - \dfrac{3}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {t - \dfrac{3}{2}} \right)^2} \ge 0\forall t > 0\\
 \to {\left( {t - \dfrac{3}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
 \to Min = \dfrac{3}{4}\\
 \Leftrightarrow t - \dfrac{3}{2} = 0\\
 \to t = \dfrac{3}{2}\\
 \to \dfrac{1}{{\sqrt x }} = \dfrac{3}{2}\\
 \to \sqrt x  = \dfrac{2}{3}\\
 \to x = \dfrac{4}{9}
\end{array}\)