Answer
$a.)$ $ \dfrac{9x-2y}{3x^2y^2}$
$b.)$ $ \dfrac{3x(x-7)}{x^2-9}$
$c.)$ $ \dfrac{x+5}{x-2}$
Work Step by Step
$a.)$ $\dfrac{3}{xy^2}-\dfrac{2}{3x^2y} = \dfrac{3(3x^2y)-2(xy^2)}{(xy^2)(3x^2y)} = \dfrac{9x^2y-2xy^2}{3x^3y^3} = \dfrac{xy(9x-2y)}{xy(3x^2y^2)} = \dfrac{9x-2y}{3x^2y^2}$
$b.)$ $\dfrac{5x}{x+3}-\dfrac{2x}{x-3} = \dfrac{5x(x-3)-2x(x+3)}{(x+3)(x-3)} = \dfrac{5x^2-15x-2x^2-6x}{(x+3)(x-3)} = \dfrac{3x^2-21x}{x^2-3^2} = \dfrac{3x(x-7)}{x^2-9}$
$c.)$ $\dfrac{x}{x-2}-\dfrac{5}{2-x} = \dfrac{x(2-x)-5(x-2)}{(x-2)(2-x)} = \dfrac{2x-x^2-5x+10}{(x-2)[-(x-2)]} = \dfrac{-x^2-3x+10}{-(x-2)^2} = \dfrac{-(x^2+3x-10)}{-(x-2)^2} = \dfrac{-(x+5)(x-2)}{-(x-2)^2} = \dfrac{-(x+5)}{-(x-2)} = \dfrac{x+5}{x-2}$
