\(A=\frac{\left(x-5\right)^2}{\left|4-x\right|}-\frac{x^2-25}{x-4}=\frac{\left(x-5\right)^2}{4-x}+\frac{x^2-25}{4-x}=\frac{\left(x-5\right)^2}{4-x}+\frac{\left(x-5\right)\left(x+5\right)}{4-x}\)
\(A=\frac{x-5}{4-x}\left(x-5+x+5\right)=\frac{2x\left(x-5\right)}{4-x}\)
\(x=3\Rightarrow A=\frac{6.\left(-2\right)}{1}=-12\)