Giải thích các bước giải: \(\begin{array}{l} m \in \left( { - \infty ; - 1} \right] \Rightarrow m \le - 1\\ y = \sqrt {(m - 3)x + m} + \sqrt {(1 - m)x + 2m - 3} \\ Dk:\left\{ \begin{array}{l} (m - 3)x + m \ge 0\\ (1 - m)x + 2m - 3 \ge 0 \end{array} \right.(*)\\ TH1:\;\left\{ \begin{array}{l} m - 3 < 0\\ 1 - m > 0 \end{array} \right. \Leftrightarrow m < 1\\ Khi\;do\;(*) \Leftrightarrow \left\{ \begin{array}{l} x \le \frac{{ - m}}{{m - 3}}\\ x \ge \frac{{2m - 3}}{{m - 1}} \end{array} \right. \Rightarrow \frac{{ - m}}{{m - 3}} \ge - 1 \Leftrightarrow \frac{{ - m}}{{m - 3}} + 1 \ge 0 \Leftrightarrow \frac{{ - 3}}{{m - 3}} \ge 0 \to t/m\;\forall m < 1\\ TH2:\left\{ \begin{array}{l} m - 3 > 0\\ 1 - m < 0 \end{array} \right. \Leftrightarrow m > 3 \to khong\;thoa\,man\\ TH3:\left\{ \begin{array}{l} m - 3 > 0\\ 1 - m > 0 \end{array} \right. \Leftrightarrow 3 < m < 1 \to khong\;thoa\;man\\ TH4:\left\{ \begin{array}{l} m - 3 < 0\\ 1 - m < 0 \end{array} \right. \Leftrightarrow 1 < m < 3\\ (*) \Leftrightarrow \left\{ \begin{array}{l} x \le \frac{{ - m}}{{m - 3}}\\ x \le \frac{{2m - 3}}{{m - 1}} \end{array} \right. \Rightarrow \left[ \begin{array}{l} \frac{{ - m}}{{m - 3}} \ge \frac{{2m - 3}}{{m - 1}} \ge - 1\\ \frac{{2m - 3}}{{m - 1}} \ge \frac{{ - m}}{{m - 3}} \ge - 1 \end{array} \right. \Rightarrow m \ge \frac{4}{3}\\ \end{array}\)