Đáp án: C
Giải thích các bước giải:
$log_3(x^2+3x+1)+log_\frac{1}{3}(\sqrt[]{3x^2+6x}+2x)=0$
$\rightarrow log_3(x^2+3x+1)-log_3(\sqrt[]{3x^2+6x}+2x)=0$
$\rightarrow log_3(x^2+3x+1)=log_3(\sqrt[]{3x^2+6x}+2x)$
$\rightarrow x^2+3x+1=\sqrt[]{3x^2+6x}+2x$
$\rightarrow x^2+x+1=\sqrt[]{3x^2+6x}$
$\rightarrow (x^2+x+1)^2=(\sqrt[]{3x^2+6x})^2$
$\rightarrow x^4+2x^3+3x^2+2x+1=3x^2+6x$
$\rightarrow \left(x-1\right)\left(x^3+3x^2+3x-1\right)=0$
$\rightarrow (x-1)((x+1)^3-2)=0$
$\rightarrow x=1\quad hoặc\quad x=-1+\sqrt[3]{2}$
$\rightarrow\begin{cases}a=1\\b=-1+\sqrt[3]{2}\end{cases}\rightarrow S=3$
