Đáp án:
Để pt có 2 nghiệm phân biệt thì:
$\begin{array}{l}
\left\{ \begin{array}{l}
m \ne 5\\
\Delta ' > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 5\\
{\left( {m - 1} \right)^2} - \left( {m - 5} \right).m > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 5\\
{m^2} - 2m + 1 - {m^2} + 5m > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 5\\
m > - \frac{1}{3}
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{ - 2m + 2}}{{m - 5}}\\
{x_1}{x_2} = \frac{m}{{m - 5}}
\end{array} \right.\\
{x_1} < 2 < {x_2}\\
\Rightarrow \left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) < 0\\
\Rightarrow {x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 4 < 0\\
\Rightarrow \frac{m}{{m - 5}} - 2.\frac{{ - 2m + 2}}{{m - 5}} + 4 < 0\\
\Rightarrow \frac{{m + 4m - 4 + 4m - 20}}{{m - 5}} < 0\\
\Rightarrow \frac{{9m - 24}}{{m - 5}} < 0\\
\Rightarrow \frac{8}{3} < m < 5\\
Vậy\,\frac{8}{3} < m < 5
\end{array}$
