\(\dfrac{1}{2x^2+3x-5}=\dfrac{1}{\left(2x+5\right)\left(x-1\right)}=\dfrac{x-3}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}\)
\(\dfrac{x+2}{4x-x^2-3}=\dfrac{-\left(x+2\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{-\left(x+2\right)\left(2x+5\right)}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}\)