ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có: \(\dfrac{B}{A}=\dfrac{\sqrt{x}}{\sqrt{x}-2}:\dfrac{\sqrt{x}+5}{2\sqrt{x}-4}\)
\(\Leftrightarrow\dfrac{B}{A}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}+5}\)
\(\Leftrightarrow\dfrac{B}{A}=\dfrac{2\sqrt{x}}{\sqrt{x}+5}\)
Để \(\dfrac{B}{A}\) nguyên thì \(2\sqrt{x}⋮\sqrt{x}+5\)
\(\Leftrightarrow2\sqrt{x}+10-10⋮\sqrt{x}+5\)
mà \(2\sqrt{x}+10⋮\sqrt{x}+5\)
nên \(-10⋮\sqrt{x}+5\)
\(\Leftrightarrow\sqrt{x}+5\inƯ\left(-10\right)\)
\(\Leftrightarrow\sqrt{x}+5\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
\(\Leftrightarrow\sqrt{x}+5\in\left\{5;10\right\}\)(Vì \(\sqrt{x}+5\ge5\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\sqrt{x}\in\left\{0;5\right\}\)
hay \(x\in\left\{0;25\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;25\right\}\)
Vậy: Để \(\dfrac{B}{A}\) nguyên thì \(x\in\left\{0;25\right\}\)