\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2+2+1}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)+3}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}-2}\)
\(=1+\dfrac{3}{\sqrt{x}-2}\)
Để A nguyên thì \(\dfrac{3}{\sqrt{x}-2}\) phải nguyên
Do đó 3⋮( \(\sqrt{x}\) -2)
⇒\(\sqrt{x}-2\) ∈ Ư(3)
Mà Ư(-1;1;-3;3)
Nên \(\left[{}\begin{matrix}\sqrt{x}-2=-1\\\sqrt{x}-2=1\\\sqrt{x}-2=-3\\\sqrt{x}-2=3\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=3\\\sqrt{x}=-1\left(v\text{ô}l\text{í}\right)\\\sqrt{x}=5\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=1\left(nh\text{ận}\right)\\x=9\left(nh\text{ận}\right)\\x=25\left(nh\text{ận}\right)\end{matrix}\right.\)
Vậy x=1 hay x=9 hay x=25 thì A nguyên
