Điều kiện: \(x \ne 0.\) Khi đó:
\(\left( {{2^{{x^2} - 4}} - 1} \right).\ln {x^2} < 0 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} {2^{{x^2} - 4}} - 1 < 0\\ \ln {x^2} > 0 \end{array} \right.\\ \left\{ \begin{array}{l} {2^{{x^2} - 4}} - 1 > 0\\ \ln {x^2} < 0 \end{array} \right. \end{array} \right.\)
TH1:
\(\begin{array}{l} \left\{ \begin{array}{l} {2^{{x^2} - 4}} - 1 < 0\\ \ln {x^2} > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {2^{{x^2} - 4}} < {2^0}\\ \ln {x^2} > \ln 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^2} - 4 < 0\\ {x^2} > 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} - 2 < x < 2\\ \left[ \begin{array}{l} x > 1\\ x < - 1 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 2 < x < - 1\\ 1 < x < 2 \end{array} \right. \end{array}\)
TH2: \(\left\{ \begin{array}{l} {2^{{x^2} - 4}} - 1 > 0\\ \ln {x^2} < 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^2} - 4 > 0\\ {x^2} < 1 \end{array} \right.\) (Loại).
