Chọn phương án B.

\(\begin{aligned}
&\,\log_6\left(3\cdot4^x+2\cdot9^x\right)=x+1\\
\Leftrightarrow&\,6^{\log_6\left(3\cdot4^x+2\cdot9^x\right)}=6^{x+1}\\
\Leftrightarrow&\,3\cdot4^x+2\cdot9^x=6\cdot6^x\\
\Leftrightarrow&\,3+2\cdot\dfrac{9^x}{4^x}=6\cdot\dfrac{6^x}{4^x}\\
\Leftrightarrow&\,3+2\cdot\left(\dfrac{3}{2}\right)^{2x}=6\cdot\left(\dfrac{3}{2}\right)^x\\
\Leftrightarrow&\,2\cdot\left(\dfrac{3}{2}\right)^{2x}-6\cdot\left(\dfrac{3}{2}\right)^x+3=0\\
\Leftrightarrow&\left[\begin{array}{l}\left(\dfrac{3}{2}\right)^x=t_1\\ \left(\dfrac{3}{2}\right)^x=t_2\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=\log_{\tfrac{3}{2}}t_1\\ x=\log_{\tfrac{3}{2}}t_2\end{array}\right.
\end{aligned}\)

Áp dụng định lý Vi-ét ta có \(t_1\cdot t_2=\dfrac{c}{a}=\dfrac{3}{2}\).

Khi đó $$\begin{aligned}
x_1+x_2&=\log_{\tfrac{3}{2}}t_1+\log_{\tfrac{3}{2}}t_2\\
&=\log_{\tfrac{3}{2}}\left(t_1\cdot t_2\right)\\
&=\log_{\tfrac{3}{2}}\dfrac{3}{2}=1.
\end{aligned}$$