Đáp án:
$\begin{array}{l}
{3^{x + 1}} + {4^{x + 1}} = {12^x} + 12\\
\Leftrightarrow {3.3^x} + {4.4^x} - {3^x}{.4^x} - 12 = 0\\
\Leftrightarrow {3^x}\left( {3 - {4^x}} \right) + 4\left( {{4^x} - 3} \right) = 0\\
\Leftrightarrow \left( {3 - {4^x}} \right)\left( {{3^x} - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{4^x} = 3\\
{3^x} = 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = {\log _4}3\\
x = {\log _3}4
\end{array} \right.\\
\Rightarrow P = {\log _4}3.{\log _3}4 = 1
\end{array}$
