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Điều kiện: $x > 1$
$\begin{array}{l}
\left( 1 \right) \Leftrightarrow {\log _2}\left( {{x^2} + 3} \right) - {\log _2}5 = - {\log _2}\left( {x
- 1} \right) - {\log _2}\left( {x + 1} \right)\\
\,\,\,\,\,\,\, \Leftrightarrow {\log _2}\left( {{x^2} + 3} \right)\left( {{x^2} - 1} \right) = {\log
_2}5\\
\,\,\,\,\,\,\, \Leftrightarrow \left( {{x^2} + 3} \right)\left( {{x^2} - 1} \right) = 5\\
\,\,\,\,\,\,\, \Leftrightarrow {x^4} + 2{x^2} - 8 = 0 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = - 4\,\,\,\,\left( {loai} \right)\\
{x^2} = 2
\end{array} \right. \left[ \begin{array}{l}
x = - \sqrt 2 \,\,\,\,\,\left( {loạ i} \right)\\
x = \sqrt 2
\end{array} \right.
\end{array}$
Vậy $x = \sqrt 2 $
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Trả lời 18-09-12 04:40 PM
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