Tìm số nghiệm của phương trình log 5 (1+x^2) + log 1/3

* Hướng dẫn giải

Đáp án B

$\begin{array}{l}PT\iff \left\{\begin{array}{l}1-x^2>0\\ {\log }_{5}3.{\log }_3\left(1+x^2\right)={\log }_3\left(1-x^2\right)\end{array}\right.\left(1\right)\\ TH1:{\log }_3\left(1+x^2\right)=0\Rightarrow x=0\Rightarrow \left(1\right)\iff \left\{\begin{array}{l}-1<x<1\\ x=0\end{array}\right.\Rightarrow x=0\\ TH2:{\log }_3\left(1+x^2\right)>0\Rightarrow x\ne 0\Rightarrow \left(1\right)\iff \left\{\begin{array}{l}-1<x<1\\ {\log }_{1+x^2}\left(1-x^2\right)={\log }_{3}5\end{array}\right.\iff \left\{\begin{array}{l}-1<x<1\\ \left\{\begin{array}{l}1-x^2=3^n\\ 1-x^2=5^n\end{array}\right.\end{array}\right.\left(2\right)\end{array}$

Vì $x\ne 0\Rightarrow \left\{\begin{array}{l}1-x^2<0\\ 1-x^2>0\end{array}\right.\Rightarrow \left(2\right)$ vô nghiệm

Kết hợp 2TH, suy ra x = 0