Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
B = m\\
\Rightarrow \frac{{\sqrt x - 2}}{{\sqrt x - 1}} = m\\
\Rightarrow \sqrt x - 2 = m.\sqrt x - m\\
\Rightarrow \left( {m - 1} \right).\sqrt x = m - 2\\
\Rightarrow \sqrt x = \frac{{m - 2}}{{m - 1}}\\
\Rightarrow \frac{{m - 2}}{{m - 1}} > 0; \ne 1\\
\Rightarrow \left\{ \begin{array}{l}
\frac{{m - 2}}{{m - 1}} > 0\\
\frac{{m - 2}}{{m - 1}} \ne 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 2\\
m < 1
\end{array} \right.\\
\frac{{m - 2 - m + 1}}{{m - 1}} \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 2\\
m < 1
\end{array} \right.\\
- 1 \ne 0\left( {tm} \right)
\end{array} \right.\\
Vậy\,m > 2/m < 1
\end{array}$
