a)
ĐKXĐ: \(x\notin\left\{0;3;-3\right\}\)
Ta có: \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{x\left(x-3\right)}\right):\left(\dfrac{-x^2}{3\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right)\)
\(=\left(\dfrac{x\left(x-3\right)}{3x\left(x-3\right)}+\dfrac{9}{3x\left(x-3\right)}\right):\left(\dfrac{-x^2}{3\left(x-3\right)\left(x+3\right)}+\dfrac{3\left(x-3\right)}{3\left(x-3\right)\left(x+3\right)}\right)\)
\(=\dfrac{x^2-3x+9}{3x\left(x-3\right)}:\dfrac{-x^2+3x-9}{3\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x+9}{3x\left(x-3\right)}\cdot\dfrac{3\left(x-3\right)\left(x+3\right)}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-x-3}{x}\)
b) Để A nguyên thì \(-x-3⋮x\)
mà \(-x⋮x\)
nên \(-3⋮x\)
\(\Leftrightarrow x\inƯ\left(-3\right)\)
\(\Leftrightarrow x\in\left\{1;-1;3;-3\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{1;-1\right\}\)
Vậy: Để A nguyên thì \(x\in\left\{1;-1\right\}\)