Đáp án:
$a) m \in \varnothing\\ b) m \ge \dfrac{2}{3}$
Giải thích các bước giải:
$a)\log_5(x^2-mx+m-2)\\ \text{ĐKXĐ:}x^2-mx+m-2 >0 \ \forall x\\ \Leftrightarrow \left\{\begin{array}{l} a>0\\ \Delta <0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 1>0\\ m^2-4(m-2) <0\end{array} \right.\\ \Leftrightarrow m^2-4x+8<0\\ \Leftrightarrow m^2-4x+4+4<0\\ \Leftrightarrow (m-2)^2+4<0\\ \Leftrightarrow m \in \varnothing\\ b)y=\sqrt{\log_3(x^2-2x+3m)}\\ \text{ĐKXĐ:} \left\{\begin{array}{l} \log_3(x^2-2x+3m) \ge 0 \ \forall x \\ x^2-2x+3m >0 \ \forall x\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x^2-2x+3m \ge 1 \ \forall x \\ x^2-2x+3m >0 \ \forall x\end{array} \right.\\ \Leftrightarrow x^2-2x+3m \ge 1\\ \Leftrightarrow x^2-2x+3m -1 \ge 0\\ \Leftrightarrow \left\{\begin{array}{l} a >0 \\ \Delta' \le 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 1 >0 \\ 1^2-3m+1 \le 0\end{array} \right.\\ \Leftrightarrow 2-3m \le 0\\ \Leftrightarrow m \ge \dfrac{2}{3}$
