Giải thích các bước giải:
\(\begin{array}{l}
1,\\
A = \dfrac{{3\sqrt x }}{{1 - \sqrt x }} = \dfrac{{3\sqrt 9 }}{{1 - \sqrt 9 }} = \dfrac{{3.3}}{{1 - 3}} = - \dfrac{9}{2}\\
2,\\
B = \dfrac{1}{{\sqrt x + 2}} - \dfrac{{x + 12}}{{4 - x}} - \dfrac{4}{{\sqrt x - 2}}\\
= \dfrac{1}{{\sqrt x + 2}} + \dfrac{{x + 12}}{{x - 4}} - \dfrac{4}{{\sqrt x - 2}}\\
= \dfrac{1}{{\sqrt x + 2}} + \dfrac{{x + 12}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} - \dfrac{4}{{\sqrt x - 2}}\\
= \dfrac{{\left( {\sqrt x - 2} \right) + x + 12 - 4\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x - 2 + x + 12 - 4\sqrt x - 8}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
3,\\
AB = \dfrac{{3\sqrt x }}{{1 - \sqrt x }}.\dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} = \dfrac{{ - 3\sqrt x }}{{\sqrt x + 2}}\\
AB > - \dfrac{3}{4}\\
\Leftrightarrow \dfrac{{ - 3\sqrt x }}{{\sqrt x + 2}} > - \dfrac{3}{4}\\
\Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x + 2}} < \dfrac{1}{4}\\
\Leftrightarrow 4\sqrt x < \sqrt x + 2\\
\Leftrightarrow 3\sqrt x < 2\\
\Leftrightarrow \sqrt x < \dfrac{2}{3}\\
\Leftrightarrow x < \dfrac{4}{9}\\
\Rightarrow 0 \le x < \dfrac{4}{9}
\end{array}\)
