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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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Đăng bài 04-05-12 12:13 PM
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