Đáp án:
$\begin{array}{l}
1)\sqrt {4 - 2\sqrt 3 } + \sqrt {4 + 2\sqrt 3 } \\
= \sqrt {3 - 2\sqrt 3 + 1} + \sqrt {3 + 2\sqrt 3 + 1} \\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= \sqrt 3 - 1 + \sqrt 3 + 1\\
= 2\sqrt 3 \\
2)a)x = 4\left( {tmdk} \right)\\
\sqrt x = 2\\
\Leftrightarrow A = \dfrac{{2\sqrt x }}{{x - 9}} = \dfrac{{2.2}}{{4 - 9}} = \dfrac{{ - 4}}{5}\\
b)Dkxd:x \ge 0;x \ne 9\\
B = \dfrac{{2\sqrt x }}{{\sqrt x - 3}} - \dfrac{5}{{\sqrt x + 3}} + \dfrac{{2x + 12}}{{9 - x}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x + 3} \right) - 5\left( {\sqrt x - 3} \right) - 2x - 12}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2x + 6\sqrt x - 5\sqrt x + 15 - 2x - 12}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{1}{{\sqrt x - 3}}\\
c)\dfrac{A}{B} < \dfrac{2}{3}\\
\Leftrightarrow \dfrac{{2\sqrt x }}{{x - 9}}:\dfrac{1}{{\sqrt x - 3}} - \dfrac{2}{3} < 0\\
\Leftrightarrow \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\left( {\sqrt x - 3} \right) - \dfrac{2}{3} < 0\\
\Leftrightarrow \dfrac{{2\sqrt x }}{{\sqrt x + 3}} - \dfrac{2}{3} < 0\\
\Leftrightarrow \dfrac{{6\sqrt x - 2\sqrt x - 6}}{{3\left( {\sqrt x + 3} \right)}} < 0\\
\Leftrightarrow 4\sqrt x - 6 < 0\\
\Leftrightarrow \sqrt x < \dfrac{3}{2}\\
\Leftrightarrow x < \dfrac{9}{4}\\
Vay\,0 \le x < \dfrac{9}{4}
\end{array}$
