Đáp án: $m < - 1\,hoặc\,1 < m < 5$
Giải thích các bước giải:
Pt có 2 nghiệm bé hơn 2 khi:
$\begin{array}{l}
\left\{ \begin{array}{l}
m - 1 \ne 0\\
\Delta ' > 0\\
{x_1} < {x_2} < 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
{\left( {m - 3} \right)^2} - \left( {m - 1} \right)\left( {m - 4} \right) > 0\\
{x_1} + {x_2} < 4\\
{x_1}{x_2} < 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
{m^2} - 6m + 9 - {m^2} + 5m - 4 > 0\\
\dfrac{{2\left( {m - 3} \right)}}{{m - 1}} < 4\\
\dfrac{{m - 4}}{{m - 1}} < 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
m < 5\\
\dfrac{{2m - 6 - 4m + 4}}{{m - 1}} < 0\\
\dfrac{{m - 4 - 4m + 4}}{{m - 1}} < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
m < 5\\
\dfrac{{ - 2m - 2}}{{m - 1}} < 0\\
\dfrac{{ - 3m}}{{m - 1}} < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 1;m < 5\\
\left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.\\
\left[ \begin{array}{l}
m > 1\\
m < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
1 < m < 5\\
m < - 1
\end{array} \right.\\
Vậy\,m < - 1\,hoặc\,1 < m < 5
\end{array}$
