Đáp án:
$\begin{array}{l}
\sqrt {\left( {1 - 2x} \right)\left( {3 - x} \right)} = 2{x^2} - 5x + 3 + m\\
Đặt:\sqrt {\left( {1 - 2x} \right)\left( {3 - x} \right)} = a\left( {a \ge 0} \right)\\
\Rightarrow {a^2} = 2{x^2} - 5x + 3\\
\Rightarrow a = {a^2} + m\\
\Rightarrow {a^2} - a + m = 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta \ge 0\\
1 > 0\\
m > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
1 - 4m \ge 0\\
m > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \le \dfrac{1}{4}\\
m > 0
\end{array} \right.\\
Vậy\,0 < m \le \dfrac{1}{4}
\end{array}$
