Đáp án:
\(x \in \left( { - \infty ;2} \right] \cup \left( {4; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
TH1:{x^2} - 8 \ge 0 \to x \in \left( { - \infty ; - 2\sqrt 2 } \right] \cup \left[ {2\sqrt 2 ; + \infty } \right)\\
Bpt \to {x^2} - 8 > 2x\\
\to {x^2} - 2x - 8 > 0\\
\to \left( {x - 4} \right)\left( {x + 2} \right) > 0\\
\to x \in \left( { - \infty ; - 2} \right) \cup \left( {4; + \infty } \right)\\
\Rightarrow x \in \left( { - \infty ; - 2\sqrt 2 } \right] \cup \left( {4; + \infty } \right)\\
TH2:{x^2} - 8 < 0 \to x \in \left( { - 2\sqrt 2 ;2\sqrt 2 } \right)\\
Bpt \to 8 - {x^2} > 2x\\
\to {x^2} + 2x - 8 < 0\\
\to \left( {x - 2} \right)\left( {x + 4} \right) < 0\\
\to x \in \left( { - 4;2} \right)\\
\Rightarrow x \in \left( { - 2\sqrt 2 ;2} \right)\\
KL:x \in \left( { - \infty ;2} \right] \cup \left( {4; + \infty } \right)
\end{array}\)
