Condition : \(x^2+8x-20=x^2+10x-2x-20=x\left(x+10\right)-2\left(x+10\right)\)

\(=\left(x-2\right)\left(x+10\right)\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2\ne0\\x+10\ne0\end{matrix}\right.\)\(\Rightarrow x\ne2;-10\)

Case 1 : \(x-2>0\Leftrightarrow x>2\)

\(A=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3=\dfrac{x}{x+10}+12x-3\)

Case 2 : \(x-2< 0\Leftrightarrow x< 2\left(x\ne-10\right)\)

\(A=\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3=\dfrac{-x}{x+10}+12x-3\)

So, \(A=\dfrac{x}{x+10}+12x-3\) when x > 2

\(A=-\dfrac{x}{x+10}+12x-3\) when \(x< 2;x\ne-10\)