Answer:
Step-by-step explanation:
[tex]\frac{6a}{a^2-6a+9}}+\frac{9}{a^2-2a-3)}\\\\=\frac{6a}{(a-3)^2}+\frac{9}{(a-3)(a+1)}\\\\\\[/tex]
[tex]=\frac{6a(a+1)+9(a-3)}{(a-3)^2(a+1)}\\\\=\frac{6a^2+6a+9a-27}{(a-3)^2(a+1)}\\[/tex]
[tex]=\frac{6a^2+15a-27}{(a-3)^2(a+1)}\\\\=\frac{3(2a^2+5a-9)}{(a-3)^2(a+1)}[/tex]
