Cho phương trình ${{\log }_{5}}\left( {{x}^{3}}-x \right)+{{\log...

Ta có \)">{{\log }_{5}}\left( {{x}^{3}}-x \right)+{{\log }_{0,2}}\left( {{x}^{2}}-2 \right)=0\Leftrightarrow {{\log }_{5}}\left( {{x}^{3}}-x \right)={{\log }_{5}}\left( {{x}^{2}}-2 \right)\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{3}}-x>0 \\
& {{x}^{3}}-x={{x}^{2}}-2 \\
\end{aligned} \right. \left\{ \begin{aligned}
& {{x}^{2}}-2>0 \\
& {{x}^{3}}-x={{x}^{2}}-2 \\
\end{aligned} \right. \left\{ \begin{aligned}
& {{x}^{3}}-x>0 \\
& {{x}^{2}}-2>0 \\
& {{x}^{3}}-x={{x}^{2}}-2. \\
\end{aligned} \right.$
Dễ thấy đáp án B sai.