Đáp án:
$\begin{array}{l}
a){x^2} - 2x + m - 1 = 0\\
\Delta ' \ge 0\\
\Leftrightarrow 1 - m + 1 \ge 0\\
\Leftrightarrow m \le 2\\
Vậy\,m \le 2\\
b)m \le 2\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
A = x_1^3 + x_2^3\\
= {\left( {{x_1} + {x_2}} \right)^3} - 3{x_1}{x_2}\left( {{x_1} + {x_2}} \right)\\
= {2^3} - 3.\left( {m - 1} \right).2\\
= 8 - 6m + 6\\
= 14 - 6m\\
Khi:m \le 2\\
\Leftrightarrow - 6m \ge - 12\\
\Leftrightarrow 14 - 6m \ge 2\\
\Leftrightarrow A \ge 2\\
\Leftrightarrow GTNN:A = 2\\
Khi:m = 2
\end{array}$
