Đáp án:
$\begin{array}{l}
2)Dkxd:x \ge 0;x \ne 1\\
B = \dfrac{{\sqrt x }}{{\sqrt x + 1}} + \dfrac{{3\sqrt x + 1}}{{x - 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}} + \dfrac{{3\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + 3\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - \sqrt x + 3\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
3)\dfrac{A}{B} = \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}.\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} = m\\
\Rightarrow m.\sqrt x + m = \sqrt x - 2\\
\Rightarrow \left( {m - 1} \right).\sqrt x = - m - 2\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
\dfrac{{ - m - 2}}{{m - 1}} \ge 0\\
\dfrac{{ - m - 2}}{{m - 1}} \ne 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
\dfrac{{m + 2}}{{m - 1}} \le 0\\
\dfrac{{ - m - 2 - m + 1}}{{m - 1}} \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
- 2 \le m < 1\\
- 2m - 1 \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
m \ne - \dfrac{1}{2}\\
- 2 \le m < 1
\end{array} \right.\\
\Rightarrow - 2 \le m < 1;m \ne - \dfrac{1}{2}
\end{array}$
