c)
\frac{2x-3}{x^2} * \frac{5x}{4x^2-9}=
x^2\ne0 =>x\ne0
4x^2-9\ne0
(2x-3)(2x+3)\ne0
2x-3\ne0=> x\ne\frac{3}{2}
2x+3\ne0=> x\ne -\frac{3}{2}
D\in R \ {0,3/2,-3/2}
\frac{2x-3}{x^2} * \frac{5x}{4x^2-9}=\frac{10x^2-15x}{x^2(4x^2-9)}=\frac{5x(2x-3)}{x^2(2x-3)(2x+3)}=\frac{5}{x(2x+3)}=\frac{5}{2x^2+3x}
d)
\frac{x^2}{x^2-4} : \frac{3x}{x+2}
x^2-4\ne0
(x-2)(x+2)\ne0
x\ne2 i x\ne-2
D\in R \ {2,-2}
\frac{x^2}{x^2-4} * \frac{x+2}{3x}=\frac{x^3+2x^2}{3x(x-2)(x+2)}=\frac{x^2(x+2)}{3x(x-2)(x+2)}=\frac{x}{3(x-2)}