\(M=\dfrac{4-x^2}{2-x}.\dfrac{x-3}{x^2-9}=\dfrac{\left(x^2-4\right).\left(x-3\right)}{\left(x-2\right).\left(x^2-9\right)}\\ =\dfrac{\left(x-2\right)\left(x+2\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\dfrac{x+2}{x+3}\\ b,x=2;G=\dfrac{x+2}{x+3}\\ Tại,x=2\Rightarrow G=\dfrac{x+2}{x+3}=\dfrac{2+2}{2+3}=\dfrac{4}{5}\)
Vậy tại x=2 thì \(G=\dfrac{4}{5}\)