Đáp án:
$\begin{array}{l}
1)Dkxd:x > 0;x\# 1\\
{x^2} = 16\\
\Leftrightarrow x = 4\left( {x > 0} \right)\\
A = \dfrac{{\sqrt x + 2}}{{x - \sqrt x }}\\
= \dfrac{{\sqrt 4 + 2}}{{4 - \sqrt 4 }} = \dfrac{4}{2} = 2\\
2)M = A:B\\
= \dfrac{{\sqrt x + 2}}{{x - \sqrt x }}:\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{{\sqrt x }}{{\sqrt x + 1}} + \dfrac{{x + 3}}{{x - 1}}} \right)\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\sqrt x + 1 - \sqrt x \left( {\sqrt x - 1} \right) + x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1 - x + \sqrt x + x + 3}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x }}.\dfrac{{\sqrt x + 1}}{{2\sqrt x + 4}}\\
= \dfrac{{\sqrt x + 1}}{{2\sqrt x }}\\
3)M = \dfrac{1}{k}\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{2\sqrt x }} = \dfrac{1}{k}\\
\Leftrightarrow k.\sqrt x + k = 2\sqrt x \\
\Leftrightarrow \left( {2 - k} \right).\sqrt x = k\\
\Leftrightarrow \left\{ \begin{array}{l}
k\# 2;k\# 0\\
\dfrac{k}{{2 - k}} > 0\\
\dfrac{k}{{2 - k}}\# 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
k\# 0;k\# 2;k\# 1\\
0 < k < 2
\end{array} \right.\\
Vậy\,0 < k < 2;k\# 1
\end{array}$
