a: TH1: x>3
\(M=\dfrac{2\left(x-3\right)}{\left(x+5\right)\left(x-3\right)}=\dfrac{2}{x+5}\)
TH2: x<3
\(M=\dfrac{-2\left(x-3\right)}{\left(x+5\right)\left(x-3\right)}=\dfrac{-2}{x+5}\)
b: TH1: x>3
Để M là số nguyên thì \(x+5\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{-1;-6;-3;-7\right\}\)
=>\(x\in\varnothing\)
TH2: x<3
Để M là số nguyên thì \(x+5\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{-4;-6;-3;-7\right\}\)