Đáp án:
$\begin{array}{l}
a)x = 2\left( {tmdk} \right)\\
\Rightarrow A = \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}} = \dfrac{{\sqrt 2 + 2}}{{\sqrt 2 + 1}}\\
= \dfrac{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 + 1}}\\
= \sqrt 2 \\
b)B = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{5\sqrt x + 2}}{{4 - x}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
c)P = - A.B\\
= - \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}.\dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
= - \dfrac{{3\sqrt x }}{{\sqrt x + 1}}\\
= - \dfrac{{3\sqrt x + 3 - 3}}{{\sqrt x + 1}}\\
= - 3 + \dfrac{3}{{\sqrt x + 1}} \in Z\\
\Rightarrow \left( {\sqrt x + 1} \right) \in \left\{ {1;3} \right\}\left( {do:\sqrt x + 1 \ge 1} \right)\\
\Rightarrow \sqrt x \in \left\{ {0;2} \right\}\\
\Rightarrow x \in \left\{ {0;4} \right\}\\
Do:x \ne 4\\
Vậy\,x = 0
\end{array}$
