Đáp án:

\(\begin{array}{l}
39)\quad S = \left\{\ln_{\tfrac{7 + 3\sqrt5}{2}}2;\ln_{\tfrac{7 + 3\sqrt5}{2}}6\right\}\\
40)\quad S = \{-2;2\}\\
41)\quad S = \{-2;2\}
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
39)\quad \left(7 + 3\sqrt5\right)^x + 12.\left(7 - 3\sqrt5\right)^x = 8.2^x\\
\Leftrightarrow \left(\dfrac{7 + 3\sqrt5}{2}\right)^x + 12.\left(\dfrac{7 - 3\sqrt5}{2}\right)^x - 8 = 0\\
\Leftrightarrow \left(\dfrac{7 + 3\sqrt5}{2}\right)^{2x}- 8.\left(\dfrac{7 + 3\sqrt5}{2}\right)^x +12= 0\\
\Leftrightarrow \left[\left(\dfrac{7 + 3\sqrt5}{2}\right)^x - 2\right]\left[\left(\dfrac{7 + 3\sqrt5}{2}\right)^x - 6\right] = 0\\
\Leftrightarrow \left[\begin{array}{l}\left(\dfrac{7 + 3\sqrt5}{2}\right)^x = 2\\\left(\dfrac{7 + 3\sqrt5}{2}\right)^x = 6\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \ln_{\tfrac{7 + 3\sqrt5}{2}}2\\x = \ln_{\tfrac{7 + 3\sqrt5}{2}}6\end{array}\right.\\
\text{Vậy}\ S = \left\{\ln_{\tfrac{7 + 3\sqrt5}{2}}2;\ln_{\tfrac{7 + 3\sqrt5}{2}}6\right\}\\
40)\quad \left(4 + \sqrt{15}\right)^x + \left(4 - \sqrt{15}\right)^x=62\\
\Leftrightarrow \left(4 + \sqrt{15}\right)^x +\dfrac{1}{\left(4 + \sqrt{15}\right)^x} - 62 =0\\
\Leftrightarrow \left(4 + \sqrt{15}\right)^{2x} - 62\left(4 + \sqrt{15}\right)^x + 1  = 0\\
\Leftrightarrow \left[\begin{array}{l}\left(4 + \sqrt{15}\right)^x =31 - 8\sqrt{15}\\\left(4 + \sqrt{15}\right)^x = 31 + 8\sqrt{15}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\left(4 + \sqrt{15}\right)^x = \left(4 + \sqrt{15}\right)^{-2}\\\left(4 + \sqrt{15}\right)^x = \left(4 + \sqrt{15}\right)^2\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = -2\\x = 2\end{array}\right.\\
\text{Vậy}\ S = \{-2;2\}\\
41)\quad \left(\sqrt{2 + \sqrt3}\right)^x +\left(\sqrt{2 -\sqrt3}\right)^x = 4\\
\Leftrightarrow \left(\sqrt{2 + \sqrt3}\right)^x + \dfrac{1}{\left(\sqrt{2 + \sqrt3}\right)^x} - 4 = 0\\
\Leftrightarrow \left(\sqrt{2 + \sqrt3}\right)^{2x} - 4\left(\sqrt{2 + \sqrt3}\right)^x + 1=0\\
\Leftrightarrow \left[\begin{array}{l}\left(\sqrt{2 + \sqrt3}\right)^x = 2 -\sqrt3\\\left(\sqrt{2 + \sqrt3}\right)^x= 2 + \sqrt3\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\left(\sqrt{2 + \sqrt3}\right)^x = \left(\sqrt{2 + \sqrt3}\right)^{-2}\\\left(\sqrt{2 + \sqrt3}\right)^x=\left(\sqrt{2 + \sqrt3}\right)^2\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = -2\\x = 2\end{array}\right.\\
\text{Vậy}\ S = \{-2;2\}
\end{array}\)