\(A=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3+5}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3}{\sqrt{x}+3}+\dfrac{5}{\sqrt{x+3}}\)

\(\Rightarrow5⋮\sqrt{x+3}\)

\(\Rightarrow\sqrt{x}+3\inƯ\left(5\right)\)

\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+3=1\Rightarrow\sqrt{x}=-2\left(KTM\right)\\\sqrt{x}+3=-1\Rightarrow\sqrt{x}=-4\left(KTM\right)\\\sqrt{x}+3=5\Rightarrow\sqrt{x}=2\Rightarrow x=4\left(TM\right)\\\sqrt{x}+3=-5\Rightarrow\sqrt{x}=-8\end{matrix}\right.\)

Vậy \(x=4\)

\(B=\sqrt{x}.A\)

\(B=\sqrt{4}.A\)

\(B=-2.A\) ( vì -2 nhỏ hơn A)

\(\Rightarrow B=-2\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)

\(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\)

\(\Rightarrow MIN_A=\dfrac{\sqrt{0}+8}{\sqrt{0}+3}=\dfrac{8}{3}\)

Vậy \(MIN_B=-2.\dfrac{8}{3}=\dfrac{-16}{3}\)