đkxđ: tự đặtpt <=> $log_x(3-x)=\frac{1}{2}.log_x(8-3x-x^2)$<=> $3-x=\sqrt{8-3x-x^2}$<=> $\begin{cases}3-x \geq 0\\ (3-x)^2=8-3x-x^2\end{cases}$
đkxđ: tự đặtpt <=> $log_x(3-x)=\frac{1}{2}.log_x(8-3x-x^2)$<=> $3-x=\sqrt{8-3x-x^2}$<=> $(3-x)^2=8-3x-x^2$
đkxđ: tự đặtpt <=> $log_x(3-x)=\frac{1}{2}.log_x(8-3x-x^2)$<=> $3-x=\sqrt{8-3x-x^2}$<=> $
\begin{cases}3-x \geq 0\\ (3-x)^2=8-3x-x^2
\end{cases}$