Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
b)x = 16\left( {tmdk} \right)\\
\Rightarrow \sqrt x = 4\\
\Rightarrow A = \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}} = \dfrac{{4 + 2}}{{4 + 1}} = \dfrac{6}{5}\\
b)B = \dfrac{{x + 12}}{{x - 4}} + \dfrac{1}{{\sqrt x + 2}} - \dfrac{4}{{\sqrt x - 2}}\\
= \dfrac{{x + 12 + \sqrt x - 2 - 4\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 12 + \sqrt x - 2 - 4\sqrt x - 8}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
\Rightarrow \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
d)A.B - 2\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}.\dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} - 2\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - 2\\
= \dfrac{{\sqrt x - 1 - 2\sqrt x - 2}}{{\sqrt x + 1}}\\
= \dfrac{{ - \sqrt x - 3}}{{\sqrt x + 1}}\\
Do:\left\{ \begin{array}{l}
- \sqrt x - 3 \le - 3 < 0\\
\sqrt x + 1 \ge 1 > 0
\end{array} \right.\\
\Rightarrow A.B - 2 < 0\\
\Rightarrow A.B < 2
\end{array}$
