Giải thích các bước giải:
⇔$\frac{1}{x-1}$- $\frac{1}{x+3}$= $\frac{2}{x+2}$- $\frac{6}{3x+5}$
⇔$\frac{x+3-x+1}{(x-1)(x+3)}$ = $\frac{6x+10-6x-18}{(3x+5)(x+3)}$
⇔$\frac{1}{(x-1)(x+3)}$= $\frac{-2}{(3x+5)(x+2)}$
⇔-2x²-2x+6=3x²+11x+10
⇔5x²+13x+4=0
Δ=13²-4×4×5=169-80=89>0
⇒√Δ=√89
⇒x=$\frac{-13±√89}{10}$