a) \(=\dfrac{3x+1+x^2-6x}{x^2-3x+1}=\dfrac{x^2-3x+1}{x^2-3x+1}=1\)
b) \(=\dfrac{6x+5x\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{6x^2+18x}{\left(x-3\right)\left(x+3\right)}=\dfrac{6x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{6x}{x-3}\)
