a, Ta có : \(A=\frac{1}{x+2}-\frac{2x}{4-x^2}+\frac{3}{x-2}\)

\(=\frac{1}{x+2}-\frac{2x}{\left(2-x\right)\left(x+2\right)}+\frac{3}{x-2}\)

\(=\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{2x}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\frac{6x+4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra : \(M=\frac{6x+4}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{3x+2}\)

\(=\frac{2\left(3x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(3x+2\right)}=\frac{2}{x-2}\)

a, \(P=\frac{3}{x+3}+\frac{1}{x-3}+\frac{18}{x^2-9}\)

\(=\frac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}=\frac{4}{x-3}\)

b, Ta có : \(\frac{4}{x-3}=2\Leftrightarrow2x-6=4\Leftrightarrow x=5\)

c, hay  \(x-3\inƯ\left(4\right)=\left\{1;2;4\right\}\)

Thử lại  --> TH1 : \(\frac{4}{4-3}=4\)( tm )

--> TH2 : \(\frac{4}{5-3}=\frac{4}{2}=2\)( tm ) 

--> TH3 : \(\frac{4}{7-3}=\frac{4}{4}=1\)( tm )

Vậy \(x=\left\{4;5;7\right\}\)thì P đạt các giá trị là số nguyên dương