Đáp án:
$\begin{array}{l}
a)m = 1\\
\Rightarrow {x^2} - 2x = 0\\
\Rightarrow x\left( {x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
Vậy\,x = 0;x = 2\\
b)x = 2\\
\Rightarrow {2^2} - 2.m.2 + m - 1 = 0\\
\Rightarrow 4 - 4m + m - 1 = 0\\
\Rightarrow 3m = 3\\
\Rightarrow m = 1\\
\Rightarrow {x_2} = 0\\
c)a.c < 0\\
\Rightarrow m - 1 < 0\\
\Rightarrow m < 1\\
Vậy\,m < 1\\
d)\Delta ' > 0\\
\Rightarrow {m^2} - m + 1 > 0\left( {tm} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
{x_1} = \frac{1}{{{x_2}}}\\
\Rightarrow {x_1}{x_2} = 1\\
\Rightarrow m - 1 = 1\\
\Rightarrow m = 2\left( {tm} \right)\\
Vậy\,m = 2\\
e)x_1^2{x_2} + {x_1}x_2^2 = 12\\
\Rightarrow {x_1}{x_2}\left( {{x_1} + {x_2}} \right) = 12\\
\Rightarrow \left( {m - 1} \right).2m = 12\\
\Rightarrow m\left( {m - 1} \right) = 6\\
\Rightarrow {m^2} - m - 6 = 0\\
\Rightarrow \left( {m - 3} \right)\left( {m + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 3\\
m = - 2
\end{array} \right.\left( {tm} \right)\\
Vậy\,m = - 2;m = 3\\
f){x_1}.{x_2} > 0\\
\Rightarrow m - 1 > 0\\
\Rightarrow m > 1\\
Vậy\,m > 1\\
g)A = x_1^2 + x_2^2\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2}\\
= {\left( {2m} \right)^2} - 2.\left( {m - 1} \right)\\
= 4{m^2} - 2m + 2\\
= 4.\left( {{m^2} - \frac{1}{2}m} \right) + 2\\
= 4.\left( {{m^2} - 2.m.\frac{1}{4} + \frac{1}{{16}}} \right) - 4.\frac{1}{{16}} + 2\\
= 4{\left( {m - \frac{1}{4}} \right)^2} - \frac{7}{4} \ge - \frac{7}{4}\\
\Rightarrow A \ge - \frac{7}{4}\\
\Rightarrow GTNN:A = - \frac{7}{4}\\
Khi:m = \frac{1}{4}
\end{array}$
