$D = (\dfrac{1}{a-\sqrt{a}} + \dfrac{1}{\sqrt{a}-1}): \dfrac{\sqrt{a}+1}{(\sqrt{a}-1)^2}$
=$\dfrac{\sqrt{a}+1}{a-\sqrt{a}}.\dfrac{(\sqrt{a}-1)^2}{\sqrt{a}+1}$
=$\dfrac{1}{\sqrt{a}(\sqrt{a}-1)}. (\sqrt{a}-1)^2$
=$\dfrac{\sqrt{a}-1}{\sqrt{a}}$
Do $\sqrt{a} \geq 0$ voi moi a nen de D>0 thi tu so lon hon 0 hay $\sqrt{a}-1>0$ hay $a>1$.