Question

Answer 1

a: \(P=\dfrac{x^2+4x-3x+12-x^2+2x}{\left(x-4\right)\left(x+4\right)}=\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x-4}\)

Answer 2

\(a,P=\dfrac{x^2+4x-3x+12-x^2+2x}{\left(x-4\right)\left(x+4\right)}=\dfrac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x-4}\\ b,P=2\Rightarrow x-4=\dfrac{3}{2}\Rightarrow x=\dfrac{11}{2}\left(tm\right)\\ c,P=\dfrac{3}{x-4}\in Z\Rightarrow x-4\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Rightarrow x\in\left\{1;3;5;7\right\}\left(tm\right)\)