a: ĐKXĐ: x<>1; x<>-1; x<>0
\(A=\dfrac{x}{3\left(x+1\right)}:\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{3\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{4x}=\dfrac{x-1}{12}\)
b: Để A là số nguyên thì x-1=12k
=>x=12k+1(k\(\in Z\))