حل المعادلة

\[\begin{array}{l}
\,\,\,\,Solve\,\,for\,\,x. \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,{\left( {9 + 4\sqrt 5 } \right)^{\frac{1}{2}x}} + \,{\left( {9 - 4\sqrt 5 } \right)^{\frac{1}{2}x}} = 18 \\
Solution:\,\,Let\,\,\,\,{\left( {9 + 4\sqrt 5 } \right)^{\frac{1}{2}x}} = u \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,{\left( {\frac{{\left( {9 + 4\sqrt 5 } \right)\left( {9 - 4\sqrt 5 } \right)}}{{\left( {9 - 4\sqrt 5 } \right)}}} \right)^{\frac{1}{2}x}} = u\, \Rightarrow \,{\left( {\frac{{81 - 80}}{{\left( {9 - 4\sqrt 5 } \right)}}} \right)^{\frac{1}{2}x}} = u \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,{\left( {\frac{1}{{\left( {9 - 4\sqrt 5 } \right)}}} \right)^{{x^2} - 3}} = u\, \Rightarrow \,{\left( {9 - 4\sqrt 5 } \right)^{{x^2} - 3}} = \frac{1}{u} \\
Hence\,\,the\,\,equation\,\,{\left( {9 + 4\sqrt 5 } \right)^{\frac{1}{2}x}} + \,{\left( {9 - 4\sqrt 5 } \right)^{\frac{1}{2}x}} = 18 \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,be:\,\,\,u + \frac{1}{u}\, = 18\, \Rightarrow \, \\
\,\,\,\,\,\,\,\, = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \frac{{18 \pm \sqrt {324 - 4} }}{2} = \frac{{18 \pm 8\sqrt 5 }}{2} = \\
{\left( {9 + 4\sqrt 5 } \right)^{\frac{1}{2}x}} = u\, \Rightarrow \,{\left( {9 + 4\sqrt 5 } \right)^{\frac{1}{2}x}} = \left( {9 \pm 4\sqrt 5 } \right)\, \Rightarrow \,\frac{1}{2}x = \pm 1\, \Rightarrow \,x = \pm 2 \\
\,\,\,\,\,\,\,\,\,\,S.S. = \left\{ { - 2,2} \right\} \\
\end{array}\]