a) \(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}.\dfrac{\sqrt{x}+2}{x+16}\)

\(B=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}.\dfrac{\sqrt{x}+2}{x+16}\)

\(B=\dfrac{\left(x+16\right).\left(\sqrt{x}+2\right)}{\left(x-16\right)\left(x+16\right)}=\dfrac{\sqrt{x}+2}{x-16}\)

b) B.(A - 1) = \(\dfrac{\sqrt{x}+2}{x-16}.\left(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}-1\right)\)

\(=\dfrac{\sqrt{x}+2}{x-16}.\dfrac{2}{\sqrt{x}+2}=\dfrac{2}{x-16}\)

B(A - 1) is interger number

<=> 2 \(⋮\) x - 16

<=> x - 16 \(\in\) {1 ; -1 ; 2 ; -2}

We have this table :

So, x = {14 ; 15 ; 17 ; 18}