Đáp án:
$\begin{array}{l}
1)a)A = \sqrt 3 + \sqrt {{{\left( {5 - \sqrt 3 } \right)}^2}} \\
= \sqrt 3 + 5 - \sqrt 3 \\
= 5\\
b)B = \left( {\dfrac{1}{{x - 2\sqrt x }} + \dfrac{3}{{\sqrt x - 2}}} \right).\dfrac{{2\sqrt x }}{{3\sqrt x + 1}}\\
= \dfrac{{1 + 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{2\sqrt x }}{{3\sqrt x + 1}}\\
= \dfrac{1}{{\sqrt x - 2}}.2\\
= \dfrac{2}{{\sqrt x - 2}}\\
2)a)m = 4\\
\Rightarrow y = \left( {4 - 2} \right).x - 4\\
\Rightarrow y = 2x - 4\\
+ Cho:x = 0 \Rightarrow y = - 4\\
+ Cho:x = 2 \Rightarrow y = 0\\
\Rightarrow \left( {0; - 4} \right);\left( {2;0} \right) \in y = 2x - 4\\
b)\left( d \right)//\left( {d'} \right)\\
\Rightarrow \left\{ \begin{array}{l}
m - 2 = - 2\\
- 4 \ne 3\left( {tm} \right)
\end{array} \right.\\
\Rightarrow m = 0
\end{array}$
