Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0\\
x = \dfrac{{2 - \sqrt 3 }}{2}\left( {tmdk} \right)\\
x = \dfrac{{4 - 2\sqrt 3 }}{4} = \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{{2^2}}}\\
\Rightarrow \sqrt x = \dfrac{{\sqrt 3 - 1}}{2}\\
P = \dfrac{{4\sqrt x }}{{x + 2\sqrt x + 1}} = \dfrac{{4\sqrt x }}{{{{\left( {\sqrt x + 1} \right)}^2}}}\\
= \dfrac{{4.\dfrac{{\sqrt 3 - 1}}{2}}}{{{{\left( {\dfrac{{\sqrt 3 - 1}}{2} + 1} \right)}^2}}} = \dfrac{{2\left( {\sqrt 3 - 1} \right)}}{{{{\left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)}^2}}}\\
= \dfrac{{2.\left( {\sqrt 3 - 1} \right).4}}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}\\
= \dfrac{{8\left( {\sqrt 3 - 1} \right).{{\left( {\sqrt 3 - 1} \right)}^2}}}{{{{\left[ {\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)} \right]}^2}}}\\
= \dfrac{{8.{{\left( {\sqrt 3 - 1} \right)}^3}}}{{{{\left( {3 - 1} \right)}^2}}}\\
= \dfrac{{8.{{\left( {\sqrt 3 - 1} \right)}^3}}}{{{2^2}}}\\
= 2.{\left( {\sqrt 3 - 1} \right)^3}
\end{array}$
