Đáp án:
$\begin{array}{l}
\left( {2 - m} \right){x^2} - 4x + {m^2} - 4m + 3 = 0\\
\Rightarrow a.c < 0\\
\Rightarrow \left( {2 - m} \right).\left( {{m^2} - 4m + 3} \right) < 0\\
\Rightarrow \left( {m - 2} \right).\left( {m - 1} \right).\left( {m - 3} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m - 2 < 0\\
\left( {m - 1} \right)\left( {m - 3} \right) < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m - 2 > 0\\
\left( {m - 1} \right)\left( {m - 3} \right) > 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m < 2\\
1 < m < 3
\end{array} \right.\\
\left\{ \begin{array}{l}
m > 2\\
\left[ \begin{array}{l}
m > 3\\
m < 1
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
1 < m < 2\\
m > 3
\end{array} \right.\\
Vậy\,1 < m < 2\,hoặc\,m > 3
\end{array}$
