\({\left( {3 + \sqrt 5 } \right)^x} + {\left( {3 - \sqrt 5 } \right)^x} = {3.2^x} \)\(\Leftrightarrow \dfrac{{{{\left( {3 + \sqrt 5 } \right)}^x}}}{{{2^x}}} + \dfrac{{{{\left( {3 - \sqrt 5 } \right)}^x}}}{{{2^x}}} = 3 \)\(\Leftrightarrow {\left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)^x} + {\left( {\dfrac{{3 - \sqrt 5 }}{2}} \right)^x} = 3.\)

Ta có $\left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)\left( {\dfrac{{3 - \sqrt 5 }}{2}} \right) = \dfrac{{{3^2} - {{\left( {\sqrt 5 } \right)}^2}}}{4} = 1 $$\Rightarrow {\left( {\dfrac{{3 - \sqrt 5 }}{2}} \right)^x} = \dfrac{1}{{{{\left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)}^x}}}$$ = {\left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)^{ - x}}.$Khi đó phương trình tương đương với \({\left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)^x} + {\left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)^{ - x}} = 3.\)

Đặt \({\left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)^x} = t\,\,\left( {t > 0} \right)\), khi đó phương trình trở thành \(t + \dfrac{1}{t} = 3 \Leftrightarrow {t^2} - 3t + 1 = 0 \)\(\Leftrightarrow \left[ \begin{array}{l}{t_1} = \dfrac{{3 + \sqrt 5 }}{2}\\{t_2} = \dfrac{{3 - \sqrt 5 }}{2}\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}{\left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)^x} = \dfrac{{3 + \sqrt 5 }}{2}\\{\left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)^x} = \dfrac{{3 - \sqrt 5 }}{2} = {\left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)^{ - 1}}\end{array} \right. \)\(\Leftrightarrow \left[ \begin{array}{l}{x_1} = 1\\{x_2} = - 1\end{array} \right. \Rightarrow {x_1}{x_2} = - 1.\)