Đáp án: C
Giải thích các bước giải:
$log_3{x^2+3x+1}+log_{\frac{1}{3}}(\sqrt[]{3x^2+6x}+2x)=0$
$\rightarrow log_3{x^2+3x+1}-log_{3}(\sqrt[]{3x^2+6x}+2x)=0$
$\rightarrow log_3{x^2+3x+1}=log_{3}(\sqrt[]{3x^2+6x}+2x)$
$\rightarrow x^2+3x+1=\sqrt[]{3x^2+6x}+2x$
$\rightarrow x^2+x+1=\sqrt[]{3x^2+6x}$
$\rightarrow (x^2+x+1)^2=3x^2+6x$
$\rightarrow x^4+x^2+1+2x^3+2x+2x^2=3x^2+6x$
$\rightarrow x^4+2x^3+3x^2+2x+1=3x^2+6x$
$\rightarrow x^4+2x^3-4x+1=0$
$\rightarrow (x^4-x^3)+(3x^3-3x)-(x-1)=0$
$\rightarrow (x-1)(x^3+3x^2+3x-1)=0$
$\rightarrow (x-1)((x+1)^3-2)$
$\rightarrow x-1=0\quad hoặc \quad (x+1)^3-2=0$
$\rightarrow x=1\quad hoặc \quad x=\sqrt[3]{2}-1$
$\rightarrow a=1, b=\sqrt[3]{2}-1$
$\rightarrow S=a^{2017}+(b+1)^3$
$\rightarrow S=1^{2017}+(\sqrt[3]{2}-1+1)^3$
$\rightarrow S=1+2$
$\rightarrow S=3$
