Đáp án:
\[\left[ \begin{array}{l}
m \ge 1\\
m \le - 1
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^2} - 2x + 1 - {m^2} \le 0,\,\,\,\,\,\,\,\forall x \in \left[ {1;2} \right]\\
\Leftrightarrow {m^2} \ge {x^2} - 2x + 1,\,\,\,\,\,\,\,\forall x \in \left[ {1;2} \right]\\
\Leftrightarrow {m^2} \ge {\left( {x - 1} \right)^2},\,\,\,\,\,\,\,\forall x \in \left[ {1;2} \right]\\
\Leftrightarrow {m^2} \ge \mathop {\max }\limits_{\left[ {1;2} \right]} f\left( x \right) = {\left( {x - 1} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
1 \le x \le 2 \Rightarrow 0 \le x - 1 \le 1 \Leftrightarrow 0 \le {\left( {x - 1} \right)^2} \le 1\\
\Rightarrow \mathop {\max }\limits_{\left[ {1;2} \right]} f\left( x \right) = f\left( 2 \right) = 1\\
\left( 1 \right) \Leftrightarrow {m^2} \ge 1\\
\Leftrightarrow \left[ \begin{array}{l}
m \ge 1\\
m \le - 1
\end{array} \right.
\end{array}\)
