ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\notin\left\{1;2\right\}\end{matrix}\right.\)
Đặt \(P=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{2\left(a-2\right)}{a+2}=\dfrac{2a-4}{a+2}\)
Để P là số nguyên thì \(2a-4⋮a+2\)
=>\(2a+4-8⋮a+2\)
=>\(-8⋮a+2\)
=>\(a+2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
=>\(a\in\left\{-1;-3;0;-4;2;-6;6;-10\right\}\)
mà a nguyên và \(\left\{{}\begin{matrix}a>0\\a\notin\left\{1;2\right\}\end{matrix}\right.\)
nên \(a=6\)
