a) \(A=\dfrac{3\left(x-3\right)+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\)
b) \(A=\dfrac{4}{x-3}=4\Leftrightarrow x-3=1\Leftrightarrow x=4\left(tm\right)\)