a: \(A=\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x+1}{x-1}\)
b: Khi x=2 thì \(A=\dfrac{2+1}{2-1}=3\)
c: Để A là số nguyên thì \(x-1+2⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{2;0;3\right\}\)