Đáp án: $\left[ \begin{array}{l}
m > 2 + 2\sqrt 2 \\
m < 2 - 2\sqrt 2
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
{\left( {m - 2} \right)^2} - 8 > 0\\
\Leftrightarrow {\left( {m - 2} \right)^2} - {\left( {2\sqrt 2 } \right)^2} > 0\\
\Leftrightarrow \left( {m - 2 - 2\sqrt 2 } \right)\left( {m - 2 + 2\sqrt 2 } \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m - 2 - 2\sqrt 2 > 0\\
m - 2 + 2\sqrt 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m - 2 - 2\sqrt 2 < 0\\
m - 2 + 2\sqrt 2 < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 2 + 2\sqrt 2 \\
m > 2 - 2\sqrt 2
\end{array} \right.\\
\left\{ \begin{array}{l}
m < 2 + 2\sqrt 2 \\
m < 2 - 2\sqrt 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
m > 2 + 2\sqrt 2 \\
m < 2 - 2\sqrt 2
\end{array} \right.\\
Vậy\,m > 2 + 2\sqrt 2 \,hoặc\,m < 2 - 2\sqrt 2
\end{array}$
