Đáp án:
$a) m \le -1 \\ b) 1 \le m \le \dfrac{4}{3} \\ c) \left[\begin{array}{l} m \ge 1 \\ m \le 0\end{array} \right.\\ d) -3 \le m \le -1 .$
Giải thích các bước giải:
$a)y=\sqrt{x-m}+\sqrt{2x-m-1}; D=(0;+\infty)\\\text{y xác định trên D}\\\Rightarrow \left\{\begin{array}{l} x-m \ge 0 \ \forall \ x >0 \\ 2x-m-1 \ge 0 \ \forall \ x >0 \end{array} \right.\\\Leftrightarrow \left\{\begin{array}{l} m \le x \ \forall \ x >0 \\ m \le 2x-1 \ \forall \ x >0 \end{array} \right.\\\Leftrightarrow \left\{\begin{array}{l} m \le 0 \\ m \le -1 \end{array} \right.\\ \Leftrightarrow m \le -1 \\b)y=\sqrt{2x-3m+4}+\dfrac{x-m}{x+m-1}; D=(0;+\infty)\\\text{y xác định trên D}\\\Rightarrow \left\{\begin{array}{l} 2x-3m+4 \ge 0 \ \forall \ x >0 \\ x+m-1 \ne 0\ \ \forall \ x >0 \end{array} \right.\\\Leftrightarrow\left\{\begin{array}{l} m \le \dfrac{2x+4}{3} \ \forall \ x >0 \\ -m \ne x-1 \end{array} \right.\\\Leftrightarrow\left\{\begin{array}{l} m \le \dfrac{4}{3} \ \forall \ x >0 \\ -m \le -1 \end{array} \right.\\\Leftrightarrow\left\{\begin{array}{l} m \le \dfrac{4}{3} \ \forall \ x >0 \\ m \ge 1 \end{array} \right.\\\Leftrightarrow 1 \le m \le \dfrac{4}{3}\\c)y=\dfrac{x+2m}{x-m+1}; D=(-1;0)\\\text{y xác định trên D}\\\Rightarrow x-m+1 \ne 0 \ \forall \ x \in (-1;0)\\\Leftrightarrow m \ne x+1 \ \forall \ x \in (-1;0)\\\Leftrightarrow \left[\begin{array}{l} m \ge 1 \\ m \le 0\end{array} \right.\\d)y=\dfrac{1}{\sqrt{x-m}}+\sqrt{-x+2m+6}; D=(-1;0)\\\text{y xác định trên D}\\\Rightarrow \left\{\begin{array}{l} x-m> 0 \ \forall \ x \in (-1;0) \\ -x+2m+6 \ge 0\ \ \forall \ x \in (-1;0) \end{array} \right.\\\Leftrightarrow \left\{\begin{array}{l} m<x \ \forall \ x \in (-1;0) \\ 2m \ge x-6\ \ \forall \ x \in (-1;0) \end{array} \right.\\\Leftrightarrow \left\{\begin{array}{l} m \le -1 \\ 2m \ge -6\end{array} \right.\\\Leftrightarrow \left\{\begin{array}{l} m \le -1 \\ m \ge -3\end{array} \right.\\\Leftrightarrow -3 \le m \le -1 .$
