Đáp án:
\(m \in \left( { - \infty ;1} \right] \cup \left\{ 2 \right\}\)
Giải thích các bước giải:
\(\eqalign{
& y = {{mx} \over {\sqrt {x - m + 2} - 1}} \cr
& DKXD:\,\,\left\{ \matrix{
x - m + 2 \ge 0 \hfill \cr
\sqrt {x - m + 2} - 1 \ne 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x \ge m - 2 \hfill \cr
x - m + 2 \ne 1 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x \ge m - 2 \hfill \cr
x \ne m - 1 \hfill \cr} \right. \cr
& \Rightarrow D = \left[ {m - 2;m - 1} \right) \cup \left( {m - 1; + \infty } \right) \cr
& Ham\,\,so\,\,xac\,\,dinh\,\,tren\,\,\left( {0;1} \right) \Leftrightarrow \left( {0;1} \right) \subset D \cr
& \Rightarrow \left[ \matrix{
m - 2 \le 0 < 1 \le m - 1 \hfill \cr
m - 1 \le 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
m \le 2 \hfill \cr
m \ge 2 \hfill \cr
m \le 1 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
m \le 1 \hfill \cr
m = 2 \hfill \cr} \right. \Leftrightarrow m \in \left( { - \infty ;1} \right] \cup \left\{ 2 \right\} \cr} \)
