a: \(P=\dfrac{2\left(x-2\right)\left(x+2\right)}{x^2+x+5}\cdot\dfrac{5\left(x^2+x+5\right)}{\left(x-4\right)\left(x+3\right)}\cdot\dfrac{\left(x-4\right)\left(x-1\right)}{10\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-1}{x+3}\)
b: Để P là số nguyên thì \(x+3-4⋮x+3\)
=>\(x+3\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{-4;-1;-5;-7\right\}\)