Lời giải:
Đặt $A=3^2+3^3+3^4+...+3^{2006}$
$\Rightarrow 3A=3^3+3^4+3^5+...+3^{2007}$
$\Rightarrow 3A-A=3^{2007}-3^2$
$\Rightarrow 2A=3^{2007}-9$
Vậy: $(4-x)+\frac{3^{2007}-9}{2}=3^{2016}:243=3^{2016}:3^5=3^{2011}$
$2(4-x)+3^{2007}-9=2.3^{2011}$
$-2x-1=2.3^{2011}-3^{2007}=3^{2007}(2.3^4-1)=161.3^{2007}$
$\Rightarrow x=\frac{1-161.3^{2007}}{2}$