Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x\# 1\\
x = \dfrac{2}{{2 + \sqrt 3 }} = \dfrac{{2.\left( {2 - \sqrt 3 } \right)}}{{{2^2} - 3}} = 4 - 2\sqrt 3 \\
= {\left( {\sqrt 3 - 1} \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 3 - 1\\
B = \dfrac{x}{{\sqrt x - 1}} = \dfrac{{4 - 2\sqrt 3 }}{{\sqrt 3 - 1 - 1}} = \dfrac{{4 - 2\sqrt 3 }}{{\sqrt 3 - 2}}\\
= \dfrac{{ - 2\left( {\sqrt 3 - 2} \right)}}{{\sqrt 3 - 2}}\\
= - 2\\
2)B = \dfrac{{x - \sqrt x + 1}}{{\sqrt x - 1}} + \dfrac{{x - 1}}{{1 - \sqrt x }}\\
= \dfrac{{x - \sqrt x + 1 - x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{2 - \sqrt x }}{{\sqrt x - 1}}\\
P = A:B\\
= \dfrac{{2 - \sqrt x }}{{\sqrt x - 1}}:\dfrac{x}{{\sqrt x - 1}}\\
= \dfrac{{2 - \sqrt x }}{{\sqrt x - 1}}.\dfrac{{\sqrt x - 1}}{x}\\
= \dfrac{{2 - \sqrt x }}{x}
\end{array}$
