Đáp án:

d. x=4

Giải thích các bước giải:

\(\begin{array}{l}
a.DK:x > 0;x \ne 1\\
P = \left( {\sqrt x  - \dfrac{1}{{\sqrt x }}} \right):\left[ {\dfrac{{\sqrt x  - 1}}{{\sqrt x }} + \dfrac{{1 - \sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}} \right]\\
 = \dfrac{{x - 1}}{{\sqrt x }}:\left[ {\dfrac{{x - 1 + 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}} \right]\\
 = \dfrac{{x - 1}}{{\sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  - 1} \right)}} = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\sqrt x }}\\
 = \dfrac{{x + 2\sqrt x  + 1}}{{\sqrt x }}\\
b.Thay:x = \dfrac{2}{{2 + \sqrt 3 }} = \dfrac{4}{{4 + 2\sqrt 3 }}\\
 = \dfrac{4}{{{{\left( {\sqrt 3  + 1} \right)}^2}}}\\
 \to P = \dfrac{{\dfrac{2}{{2 + \sqrt 3 }} + 2\sqrt {\dfrac{4}{{{{\left( {\sqrt 3  + 1} \right)}^2}}}}  + 1}}{{\sqrt {\dfrac{4}{{{{\left( {\sqrt 3  + 1} \right)}^2}}}} }}\\
 = \dfrac{{\dfrac{2}{{2 + \sqrt 3 }} + 2.\dfrac{2}{{\sqrt 3  + 1}} + 1}}{{\dfrac{2}{{\sqrt 3  + 1}}}} = \dfrac{{3 + 3\sqrt 3 }}{2}\\
c.P > 2\\
 \to \dfrac{{x + 2\sqrt x  + 1}}{{\sqrt x }} > 2\\
 \to \dfrac{{x + 2\sqrt x  + 1 - 2\sqrt x }}{{\sqrt x }} > 0\\
 \to x + 1 > 0\left( {do:\sqrt x  > 0\forall x > 0} \right)\\
 \to x >  - 1\left( {ld} \right)\forall x > 0\\
 \to dpcm\\
d.P\sqrt x  = 6\sqrt x  - 3 - \sqrt {x - 4} \\
 \to \dfrac{{x + 2\sqrt x  + 1}}{{\sqrt x }}.\sqrt x  = 6\sqrt x  - 3 - \sqrt {x - 4} \\
 \to x + 2\sqrt x  + 1 = 6\sqrt x  - 3 - \sqrt {x - 4} \\
 \to x - 4\sqrt x  + 4 =  - \sqrt {x - 4} \\
 \to {\left( {\sqrt x  - 2} \right)^2} =  - \sqrt {x - 4} \left( 1 \right)\\
Do:{\left( {\sqrt x  - 2} \right)^2} \ge 0\forall x > 0\\
\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
\sqrt x  - 2 = 0\\
 - \sqrt {x - 4}  = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
x = 4
\end{array} \right.\left( {TM} \right)\\
 \to x = 4
\end{array}\)