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TXĐ: R.
Đặt thừa số, ta có:
$\begin{array}{l}
{x^2}\left( {{2^{{x^2}}} - 4} \right) - 2x\left( {{2^{{x^2}}} - 4} \right) - 3\left( {{2^{{x^2}}} - 4} \right) < 0\\
\Leftrightarrow \left( {{2^{{x^2}}} - 4} \right)\left( {{x^2} - 2x - 3} \right) < 0
\end{array}$
$\Leftrightarrow \left\{ \begin{array}{l}
\left( {{x^2} - 2x - 3} \right) > 0\\
\left( {{2^{{x^2}}} - 4} \right) < 0
\end{array} \right.$ $(1)$ hoặc $ \left\{ \begin{array}{l}
\left( {{x^2} - 2x - 3} \right) < 0\\
\left( {{2^{{x^2}}} - 4} \right) > 0
\end{array} \right.$ $(2)$
$\begin{array}{l}
(1) \Leftrightarrow \left\{ \begin{array}{l}
- 1 < x < 3\\
{2^{{x^2}}} > 2
\end{array} \right.\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
- 1 < x < 3\\
x < - \sqrt 2 \,\,\,\,\,\, \vee \,\,\,\,\,x > \sqrt 2
\end{array} \right.\\
\,\,\,\,\,\,\,\,\, \Leftrightarrow \sqrt 2 < x < 3
\end{array}$
$\begin{array}{l}
(2) \Leftrightarrow \left\{ \begin{array}{l}
x < - 1\,\,\,\,\,\, \vee \,\,\,\,\,\,\,\,x > 3\\
{2^{{x^2}}} < 2
\end{array} \right.\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
x < - 1\,\,\,\,\,\, \vee \,\,\,\,\,\,\,\,x > 3\\
- \sqrt 2 < x < \sqrt 2 \,\,\,\,\,\,\,\,\,
\end{array} \right.\\
\,\,\,\,\,\,\,\,\,\, \Leftrightarrow - \sqrt 2 < x < - 1
\end{array}$
Vậy BPT đã cho có nghiệm
$\left[\begin{array}{l}\sqrt 2 < x < 3\\- \sqrt 2 < x < - 1\end{array} \right.$
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Đăng bài 26-04-12 09:18 AM
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