Đáp án:
\(m \le - 1\)
Giải thích các bước giải:
\(\begin{array}{l} y = \sqrt {x - m} + \sqrt {2x - m - 1} \\ DKXD:\,\,\left\{ \begin{array}{l} x - m \ge 0\\ 2x - m - 1 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge m\\ x \ge \frac{{m + 1}}{2} \end{array} \right.\,\,\left( * \right)\\ TH1:\,\,m \ge \frac{{m + 1}}{2} \Leftrightarrow 2m \ge m + 1 \Leftrightarrow m \ge 1\\ \Rightarrow \left( * \right) \Leftrightarrow x \ge m \Rightarrow D = \left[ {m; + \infty } \right)\\ De\,ham\,so\,\,xac\,\,dinh\,\,tren\,\,\left( {0; + \infty } \right)\\ \Rightarrow \left( {0; + \infty } \right) \subset \left[ {m; + \infty } \right) \Rightarrow 0 \ge m\\ \Rightarrow \left\{ \begin{array}{l} m \le 0\\ m \ge 1 \end{array} \right. \Rightarrow m \in \emptyset \\ TH2:\,\,m < \frac{{m + 1}}{2} \Leftrightarrow m < 1\\ \left( * \right) \Leftrightarrow x \ge \frac{{m + 1}}{2} \Rightarrow D = \left[ {\frac{{m + 1}}{2}; + \infty } \right)\\ De\,ham\,so\,\,xac\,\,dinh\,\,tren\,\,\left( {0; + \infty } \right)\\ \Rightarrow \left( {0; + \infty } \right) \subset \left[ {\frac{{m + 1}}{2}; + \infty } \right) \Rightarrow 0 \ge \frac{{m + 1}}{2}\\ \Leftrightarrow m + 1 \le 0 \Leftrightarrow m \le - 1\\ \Rightarrow \left\{ \begin{array}{l} m \le - 1\\ m < 1 \end{array} \right. \Leftrightarrow m \le - 1\\ Vay\,\,m \le - 1. \end{array}\)
