Решение:
\(12^{x} - 8^{x} - 2 \cdot 6^{x + 1} + 3 \cdot 4^{x + 1} + 32 \cdot 3^{x} - 2^{x + 5} \leq 0\)
\(6^{x} \cdot 2^{x} - 4^{x} \cdot 2^{x} - 2 \cdot 6^{x} \cdot 6 +\)\(3 \cdot 4^{x} \cdot 4 + 32 \cdot 3^{x} - 2^{x} \cdot 32 \leq 0\)
\(3^{x} \cdot 2^{2 x} - 2^{3 x} - 12 \cdot 6^{x} + 12 \cdot 4^{x} + 32 \cdot 3^{x} - 32 \cdot 2^{x} \leq 0\)
\(3^{x} \cdot 2^{2 x} - 2^{3 x} - 12 \cdot 2^{x} \cdot 3^{x} + 12 \cdot 2^{2 x} + 32 \cdot 3^{x} - 32 \cdot 2^{x} \leq 0\)
\(2^{2 x} \left(3^{x} - 2^{x}\right) - 2^{x} \cdot 12 \left(3^{x} - 2^{x}\right) + 32 \left(3^{x} - 2^{x}\right) \leq 0\)
\(\left(3^{x} - 2^{x}\right) \left(2^{2 x} - 2^{x} \cdot 12 + 32\right) \leq 0\)
\(\left(3^{x} - 2^{x}\right) \left(2^{x} - 4\right) \left(2^{x} - 8\right) \leq 0\)
\(\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}\left(3^{x} - 2^{x}\right) \left(2^{x} - 4\right) \leq 0 \\ 2^{x} - 8 \geq 0\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}\left(3^{x} - 2^{x}\right) \left(2^{x} - 4\right) \geq 0 \\ 2^{x} - 8 \leq 0\end{matrix} \end{matrix}\right. \end{matrix}\right.\)
\(\left(3^{x} - 2^{x}\right) \left(2^{x} - 4\right) \leq 0\)
\(\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}3^{x} - 2^{x} \leq 0 \\ 2^{x} - 4 \geq 0\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}3^{x} - 2^{x} \geq 0 \\ 2^{x} - 4 \leq 0\end{matrix} \end{matrix}\right. \end{matrix}\right.\)
\(\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}\left(\frac{3}{2}\right)^{x} \leq 1 \\ 2^{x} \geq 2^{2}\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}\left(\frac{3}{2}\right)^{x} \geq 1 \\ 2^{x} \leq 2^{2}\end{matrix} \end{matrix}\right. \end{matrix}\right.\)
\(\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}x \leq 0 \\ x \geq 2\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}x \geq 0 \\ x \leq 2\end{matrix} \end{matrix}\right. \end{matrix}\right.\)
\(x \in \left[0 ; 2\right] .\)
\(2^{x} - 8 \geq 0 \\ 2^{x} \geq 2^{3} \\ x \geq 3 .\)
\(\left(3^{x} - 2^{x}\right) \left(2^{x} - 4\right) \geq 0\)
\(\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}3^{x} - 2^{x} \geq 0 \\ 2^{x} - 4 \geq 0\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}3^{x} - 2^{x} \leq 0 \\ 2^{x} - 4 \leq 0\end{matrix} \end{matrix}\right. \end{matrix}\right.\)
\(\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}x \geq 0 \\ x \geq 2\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}x \leq 0 \\ x \leq 2\end{matrix} \end{matrix}\right. \end{matrix}\right.\)
\(x \in ( - \infty ; 0 \left]\right. \cup [ 2 ; + \infty ) .\)
\(2^{x} - 8 \leq 0 \\ x \leq 3\)
\(\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}x \in \left[0 ; 2\right] \\ x \geq 3\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}x \in ( - \infty ; 0 \left]\right. \cup [ 2 ; + \infty ) \\ x \leq 3\end{matrix} \end{matrix}\right. \end{matrix}\right.\)
\(\left[\begin{matrix} \emptyset \\ x \in ( - \infty ; 0 \left]\right. \cup [ 2 ; 3 \left]\right. . \end{matrix}\right.\)

Ответ: \(( - \infty ; 0 \left]\right. \cup [ 2 ; 3 \left]\right.\)

Источник: ФИПИ