ĐKXĐ: \(a>0;a\ne1;a\ne2\)
\(A=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\dfrac{a-2}{a+2}\)
\(A=\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\dfrac{a-2}{a+2}\)
\(A=\dfrac{2\sqrt{a}}{\sqrt{a}}\left(\dfrac{a-2}{a+2}\right)=\dfrac{2\left(a-2\right)}{a+2}\)
c/ \(A=2-\dfrac{8}{a+2}\)
Để A nguyên \(\Rightarrow8⋮\left(a+2\right)\Rightarrow a+2=Ư\left(8\right)\)
Mà \(a>0\Rightarrow a+2>2\Rightarrow\left[{}\begin{matrix}a+2=4\\a+2=8\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=2\left(l\right)\\a=6\end{matrix}\right.\)
Vậy \(a=6\) thì A nguyên
