ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{9{x^2} + 5 > 0,\,}\\{8{x^4} + 14 > 0}\end{array}\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,x\, \in \,R.} \right.\)
\(1 + {\log _2}\left( {9{x^2} + 5} \right) = {\log _{\sqrt 2 }}\sqrt {8{x^4} + 14} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _2}2 + {\log _2}\left( {9{x^2} + 5} \right) = {\log _2}\left( {8{x^4} + 14} \right)\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\log _2}\left( {18{x^2} + 10} \right) = {\log _2}\left( {8{x^4} + 14} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,18{x^2} + 10 = 8{x^4} + 14.\)
Пусть \({x^2} = t\), тогда:
\(8{t^2}-18t + 4 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 2,\,}\\{t = \dfrac{1}{4}.}\end{array}} \right.\)
Возвращаясь к прежней неизвестной, получим:
\(\left[ {\begin{array}{*{20}{c}}{{x^2} = 2,}\\{{x^2} = \dfrac{1}{4}}\end{array}\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pm \sqrt 2 ,}\\{x = \pm \dfrac{1}{2}.}\end{array}} \right.} \right.\)
Следовательно, произведение корней уравнения равно: \(\sqrt 2 \cdot \left( {-\sqrt 2 } \right) \cdot \dfrac{1}{2} \cdot \left( {-\dfrac{1}{2}} \right) = 0,5.\)
Ответ: 0,5.