\(a,=\dfrac{x^2+3+2x-3}{2xy^2}=\dfrac{x\left(x+2\right)}{2xy^2}=\dfrac{x+2}{2y^2}\\ b,=\dfrac{3x-13-2}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\\ c,=\dfrac{x^2-3x+x+3-x^2+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2}{x+3}\)


