Đáp án:
$\begin{array}{l}
1)x = \sqrt {4 + 2\sqrt 3 } - \sqrt 3 \left( {tmdk} \right)\\
= \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \sqrt 3 \\
= \sqrt 3 + 1 - \sqrt 3 \\
= 1\\
\Rightarrow \sqrt x = 1\\
\Rightarrow A = \dfrac{{1 - 3}}{1} = - 2\\
2)B = \dfrac{{2x - 2\sqrt x }}{{x - 4}} + \dfrac{1}{{\sqrt x + 2}} + \dfrac{{\sqrt x - 1}}{{2 - \sqrt x }}\\
= \dfrac{{2x - 2\sqrt x + \sqrt x - 2 - \left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{2x - \sqrt x - 2 - x - \sqrt x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 2\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 2}}\\
3)A.B = m\\
\Rightarrow \dfrac{{\sqrt x - 3}}{{\sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x + 2}} = m\\
\Rightarrow \dfrac{{\sqrt x - 3}}{{\sqrt x + 2}} = m\\
\Rightarrow m = \dfrac{{\sqrt x + 2 - 5}}{{\sqrt x + 2}} = 1 - \dfrac{5}{{\sqrt x + 2}}\\
+ )Do:\sqrt x > 0\\
\Rightarrow \sqrt x + 2 > 2\\
\Rightarrow \dfrac{1}{{\sqrt x + 2}} < \dfrac{1}{2}\\
\Rightarrow \dfrac{5}{{\sqrt x + 2}} < \dfrac{5}{2}\\
\Rightarrow 1 - \dfrac{5}{{\sqrt x + 2}} > 1 - \dfrac{5}{2} = - \dfrac{3}{2}\\
\Rightarrow m > - \dfrac{3}{2}\\
+ )Do:x \ne 4\\
\Rightarrow \sqrt x \ne 2\\
\Rightarrow m = \dfrac{{\sqrt x - 3}}{{\sqrt x + 2}} \ne \dfrac{{2 - 3}}{{2 + 2}} = \dfrac{{ - 1}}{4}\\
Vay\,m > \dfrac{{ - 3}}{2};m \ne - \dfrac{1}{4}
\end{array}$
