Giải thích các bước giải:
đkxđ : $x\ne 2,-3,-1,\dfrac12$
$\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=-\dfrac{3}{2x-1}$
$\to \dfrac{5}{(x+3)(x-2)}-\dfrac{2}{(x+1)(x+3)}=-\dfrac{3}{2x-1}$
$\to\dfrac{5}{\left(x+3\right)\left(x-2\right)}\left(x+3\right)\left(x-2\right)\left(x+1\right)\left(2x-1\right)-\dfrac{2}{\left(x+1\right)\left(x+3\right)}\left(x+3\right)\left(x-2\right)\left(x+1\right)\left(2x-1\right)=-\dfrac{3}{2x-1}\left(x+3\right)\left(x-2\right)\left(x+1\right)\left(2x-1\right)$
$\to 5\left(x+1\right)\left(2x-1\right)-2\left(x-2\right)\left(2x-1\right)=-3\left(x+3\right)\left(x-2\right)\left(x+1\right)$
$\to 6x^2+15x-9=-3x^3-6x^2+15x+18$
$\to 3x^3+12x^2-27=0$
$\to 3\left(x+3\right)\left(x^2+x-3\right)=0$
$\to x=-3$
Hoặc $x^2+x-3=0\to \:x=\dfrac{-1+\sqrt{13}}{2},\:x=\dfrac{-1-\sqrt{13}}{2}$
Vì $x\ne -3\to \:x=\dfrac{-1+\sqrt{13}}{2},\:x=\dfrac{-1-\sqrt{13}}{2}$
