Đáp án:
b) \(\left[ \begin{array}{l}
m > 5\\
- \dfrac{1}{3} < m < \dfrac{{24}}{9}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\left\{ \begin{array}{l}
m \ne 1\\
{m^2} - 4m + 4 - \left( {m - 1} \right)\left( {m - 3} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 1\\
{m^2} - 4m + 4 - {m^2} + 4m - 3 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 1\\
1 > 0\left( {ld} \right)
\end{array} \right.\\
Có:{x_1} + {x_2} + {x_1}{x_2} < 1\\
\to \dfrac{{2m - 4}}{{m - 1}} + \dfrac{{m - 3}}{{m - 1}} < 1\\
\to \dfrac{{3m - 7 - m + 1}}{{m - 1}} < 0\\
\to \dfrac{{2m - 6}}{{m - 1}} < 0\\
\to 1 < m < 3\\
b)DK:\left\{ \begin{array}{l}
m \ne 5\\
{m^2} - 2m + 1 - m\left( {m - 5} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 5\\
3m + 1 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 5\\
m > - \dfrac{1}{3}
\end{array} \right.\\
Có:{x_1} < {x_2} < 2\\
\to \left\{ \begin{array}{l}
{x_1} - 2 < 0\\
{x_2} - 2 < 0
\end{array} \right.\\
\to \left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) > 0\\
\to {x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 4 > 0\\
\to \dfrac{m}{{m - 5}} - 2.\dfrac{{ - 2m + 2}}{{m - 5}} + 4 > 0\\
\to \dfrac{{4m - 4 + m + 4m - 20}}{{m - 5}} > 0\\
\to \dfrac{{9m - 24}}{{m - 5}} > 0\\
\to \left[ \begin{array}{l}
m > 5\\
m < \dfrac{{24}}{9}
\end{array} \right.\\
KL:\left[ \begin{array}{l}
m > 5\\
- \dfrac{1}{3} < m < \dfrac{{24}}{9}
\end{array} \right.
\end{array}\)
