Question

Answer 1

\(B=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}+1}=1\)

Answer 2

\(B=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-\sqrt{a}}\)

\(\Rightarrow B=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}\)

\(\Rightarrow B=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}\)

\(\Rightarrow B=1\)