Đáp án: x=0 hoặc x=2
Giải thích các bước giải:
$\begin{array}{l}
Do:\left( {3 - \sqrt 5 } \right).\left( {3 + \sqrt 5 } \right) = 9 - 5 = 4\\
\Rightarrow {\left( {3 - \sqrt 5 } \right)^{2x - {x^2}}} + {\left( {3 + \sqrt 5 } \right)^{2x - {x^2}}} - {2^{1 + 2x - {x^2}}}\\
= {\left( {3 - \sqrt 5 } \right)^{2x - {x^2}}} + {\left( {3 + \sqrt 5 } \right)^{2x - {x^2}}} - 2.{\left( {{4^{\frac{1}{2}}}} \right)^{2x - {x^2}}}\\
= {\left( {3 - \sqrt 5 } \right)^{2x - {x^2}}} + {\left( {3 + \sqrt 5 } \right)^{2x - {x^2}}} - 2.{\left( {3 - \sqrt 5 } \right)^{\frac{{2x - {x^2}}}{2}}}.{\left( {3 + \sqrt 5 } \right)^{\frac{{2x - {x^2}}}{2}}}\\
= {\left( {{{\left( {3 - \sqrt 5 } \right)}^{\frac{{2x - {x^2}}}{2}}} - {{\left( {3 + \sqrt 5 } \right)}^{\frac{{2x - {x^2}}}{2}}}} \right)^2} \le 0\\
\Rightarrow {\left( {3 - \sqrt 5 } \right)^{\frac{{2x - {x^2}}}{2}}} = {\left( {3 + \sqrt 5 } \right)^{\frac{{2x - {x^2}}}{2}}}\\
\Rightarrow \frac{{2x - {x^2}}}{2} = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.
\end{array}$
