a, ĐKXĐ:\(5x^2-45\ne0\Rightarrow x^2-9\ne0\Rightarrow x\ne\pm3\)
b, \(\dfrac{x^2-6x+9}{5x^2-45}=\dfrac{\left(x-3\right)^2}{5\left(x^2-9\right)}=\dfrac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{5\left(x+3\right)}=\dfrac{x-3}{5x+15}\)
\(c,A=-\dfrac{2}{3}\Rightarrow\dfrac{x-3}{5x+15}=\dfrac{-2}{3}\\ \Rightarrow-2\left(5x+15\right)=3\left(x-3\right)\\ \Rightarrow-10x-30=3x-9\\ \Rightarrow3x-9+10x+30=0\\ \Rightarrow13x+39=0\\ \Rightarrow13x=-39\\ \Rightarrow x=-3\)