b) $\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}$ ($ĐKXĐ : x \neq 3,-3$ )
$\to 5.(x+3)+4.(x-3)=x-5$
$\to 9x +3=x-5$
$\to 8x = -8$
$\to x= -1$ ( Thỏa mãn )
c) $|2x-4| = 3-3x$
$\to \left[ \begin{array}{l}2x-4=3-3x\\2x-4=3x-3\end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{7}{5}\\x=1\end{array} \right.$