a: ĐKXĐ: a>=0; a<>1; a<>4
b: \(A=\dfrac{\sqrt{a}+1+\sqrt{a}-1}{a-1}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{2\sqrt{a}}{a-1}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{2\sqrt{a}\left(\sqrt{a}-2\right)}{3\left(\sqrt{a}+1\right)}\)
c: Để A>0 thì căn a-2>0
=>a>4