Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
1)x = 2\left( {tmdk} \right)\\
\Rightarrow B = \dfrac{2}{{\sqrt 2 - 1}}\\
= \dfrac{{2\left( {\sqrt 2 + 1} \right)}}{{2 - 1}}\\
= 2\sqrt 2 + 2\\
2)A = \dfrac{{x - 1}}{{\sqrt x - 1}} - \dfrac{{x\sqrt x + 1}}{{x - 1}}\\
= \dfrac{{x - 1}}{{\sqrt x - 1}} - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - 1}}{{\sqrt x - 1}} - \dfrac{{x - \sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{x - 1 - x + \sqrt x - 1}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}\\
P = \dfrac{A}{B} = \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}:\dfrac{x}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x - 2}}{x}\left( {x \ne 0} \right)\\
3)Dkxd:x > 0;x \ne 1\\
P < - 1\\
\Rightarrow \dfrac{{\sqrt x - 2}}{x} < - 1\\
\Rightarrow \dfrac{{\sqrt x - 2 + x}}{x} < 0\\
\Rightarrow \dfrac{{x + \sqrt x - 2}}{x} < 0\\
\Rightarrow x + \sqrt x - 2 < 0\left( {do:x > 0} \right)\\
\Rightarrow \left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) < 0\\
\Rightarrow \sqrt x - 1 < 0\\
\Rightarrow \sqrt x < 1\\
\Rightarrow x < 1\\
Vay\,0 < x < 1
\end{array}$
