a: \(A=\left(\dfrac{x-4}{x^3-1}-\dfrac{1}{1-x}\right):\left(1+\dfrac{8-x}{x^2+x+1}\right)\)
\(=\left(\dfrac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right):\dfrac{x^2+x+1+8-x}{x^2+x+1}\)
\(=\dfrac{x-4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x^2+9}\)
\(=\dfrac{x^2+2x-3}{\left(x-1\right)\left(x^2+9\right)}=\dfrac{x+3}{x^2+9}\)
b: Để A nguyên thì \(x+3⋮x^2+9\)
=>\(\left(x+3\right)\left(x-3\right)⋮x^2+9\)
=>\(x^2-9⋮x^2+9\)
=>\(x^2+9-18⋮x^2+9\)
=>\(-18⋮x^2+9\)
=>\(x^2+9\in\left\{9;18\right\}\)(do \(x^2+9>=9\forall x\) thỏa mãn ĐKXĐ)
=>\(x^2\in\left\{0;9\right\}\)
=>\(x\in\left\{0;3;-3\right\}\)