Đáp án:
$\begin{array}{l}
9)a)\\
+ Khi:m = 1 \Rightarrow x - 1 = 0\\
\Rightarrow x = 1\\
+ Khi:m \ne 1\\
\Rightarrow \Delta = {\left( {2m - 1} \right)^2} - 4\left( {m - 1} \right)\left( {m - 2} \right)\\
= 4{m^2} - 4m + 1 - 4{m^2} + 12m - 8\\
= 8m - 7\\
+ )\Delta \ge 0 \Rightarrow m \ge \frac{7}{8} \Rightarrow \text{pt có nghiệm}\\
+ )\Delta < 0 \Rightarrow m > \frac{7}{8} \Rightarrow \text{pt vô nghiệm}\\
b)\left\{ \begin{array}{l}
m \ne 1\\
\Delta > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
m > \frac{7}{8}
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{1 - 2m}}{{m - 1}}\\
{x_1}{x_2} = \frac{{m - 2}}{{m - 1}}
\end{array} \right.\\
3{x_1} + 3{x_2} - 4{x_1}{x_2} = 1\\
\Rightarrow 3\left( {{x_1} + {x_2}} \right) - 4{x_1}{x_2} = 1\\
\Rightarrow 3.\frac{{1 - 2m}}{{m - 1}} - 4.\frac{{m - 2}}{{m - 1}} = 1\\
\Rightarrow 3 - 6m - 4m + 8 = m - 1\\
\Rightarrow 11m = 12\\
\Rightarrow m = \frac{{12}}{{11}}\left( {tm} \right)\\
10)\\
\Delta ' > 0\\
\Rightarrow {\left( {m + 1} \right)^2} - {m^2} - 2 > 0\\
\Rightarrow 2m + 1 - 2 > 0\\
\Rightarrow m > \frac{1}{2}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = {m^2} + 2
\end{array} \right.\\
P = {x_1}.{x_2} - 2\left( {{x_1} + {x_2}} \right) - 6\\
= {m^2} + 2 - 4\left( {m + 1} \right) - 6\\
= {m^2} + 2 - 4m - 4 - 6\\
= {m^2} - 4m - 8\\
= {\left( {m - 2} \right)^2} - 12 \ge - 12\\
\Rightarrow P \ge - 12\\
\Rightarrow GTNN:P = - 12\,khi:m = 2\left( {tmdk} \right)
\end{array}$
