a: ĐKXĐ: a>=0; a<>1
b: \(C=\left[1:\dfrac{1+\sqrt{a}-\sqrt{a}}{\sqrt{a}+1}\right]\cdot\left[\dfrac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\left(a+1\right)\left(\sqrt{a}-1\right)}=\dfrac{a-1}{a+1}\)
c: Để C nguyên thì a+1-2 chia hết cho a+1
=>\(a+1\in\left\{1;-1;2;-2\right\}\)
=>\(a\in\left\{0\right\}\)