Answer:
[tex] \boxed{\boxed{\huge \sf D. \ \frac{5x - 12}{(x + 3)(x - 3)}}} [/tex]
Step-by-step explanation:
[tex] \sf Summation \: of \: following : \\ \sf \implies \frac{3}{ {x}^{2} - 9} + \frac{5}{x + 3} \\ \\ \sf {x}^{2} - 9 = {x}^{2} - {3}^{2} : \\ \sf \implies \frac{3}{ {x}^{2} - \boxed{ \sf {3}^{2} }} + \frac{5}{x + 3} \\ \\ \sf Factor \: the \: difference \: of \: two \: squares. \\ \sf {x}^{2} - {3}^{2} = (x + 3)(x - 3) : \\ \sf \implies \frac{3}{ \boxed{ \sf (x + 3)(x - 3)}} + \frac{5}{x + 3} \\ \\ \sf Put \: each \: term \: in \: \frac{3}{(x + 3)(x - 3)} + \frac{5}{x + 3} over \\ \sf the \: common \: denominator \: (x + 3)(x - 3) : \\ \sf \implies \frac{3}{(x + 3)(x - 3)} + \frac{5(x - 3)}{(x + 3)(x - 3)} \\ \\ \sf \frac{3}{(x + 3)(x - 3)} + \frac{5(x - 3)}{(x + 3)(x - 3)} = \frac{5(x - 3) + 3}{(x + 3)(x - 3)} : \\ \sf \implies \frac{5(x - 3) + 3}{(x + 3)(x - 3)}[/tex]
[tex] \sf 5(x - 3) = 5x - 15 : \\ \sf \implies \frac{ \boxed{ \sf 5x - 15} + 3}{(x + 3)(x - 3)} \\ \\ \sf Grouping \: like \: terms, \: 5x - 15 + 3 = 5x + (3 - 15) : \\ \sf \implies \frac{ \boxed{ \sf 5x + (3 - 15)}}{(x + 3)(x - 3)} \\ \\ \sf 3 - 15 = - 12 : \\ \sf \implies \frac{5x - 12}{(x + 3)(x - 3)} [/tex]
