3: \(=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
\(a,=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\\ b,=\dfrac{6x-18-6x-18+9x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{9\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{9}{x+3}\)
\(3,\dfrac{x}{x+3}+\dfrac{2x}{x-3}+\dfrac{-3x^2-9}{x^2-9}=\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{2x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{3x^2+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3}{x+3}\)
\(4,\dfrac{6}{x+3}-\dfrac{6}{x-3}+\dfrac{9x+9}{x^2-9}=\dfrac{6\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{6\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{9x+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{6x-18-6x-18+9x+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{9x-27}{\left(x+3\right)\left(x-3\right)}=\dfrac{9\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{9}{x+3}\)
