a: ĐKXĐ: a>=0; a<>1
\(A=\dfrac{a+\sqrt{a}-2-a+\sqrt{a}+2}{a-1}\cdot\dfrac{\sqrt{a}+1}{1}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}-1}\)
b: Để A là số nguyên thì \(2\sqrt{a}-2+2⋮\sqrt{a}-1\)
=>\(\sqrt{a}-1\in\left\{1;-1;2;-2\right\}\)
hay \(a\in\left\{4;0;9\right\}\)
c: \(A-2=\dfrac{2\sqrt{a}-2\sqrt{a}+2}{\sqrt{a}-1}>0\)
=>A>2