Question

\(\left(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{a-4}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+2}{\sqrt{a}}\)

Answer 1

\(=\left[\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\right]\cdot\dfrac{\sqrt{a}}{\sqrt{a}+2}\left(a>0;a\ne4\right)\\ =\left(\sqrt{a}+2+\sqrt{a}+2\right)\cdot\dfrac{\sqrt{a}}{\sqrt{a}+2}\\ =\dfrac{2\sqrt{a}\left(\sqrt{a}+2\right)}{\sqrt{a}+2}=2\sqrt{a}\)

Answer 2

=(\(\dfrac{\left[\sqrt{a}+2\right]^2}{\sqrt{a}+2}+\dfrac{\left[\sqrt{a}-2\right]\left[\sqrt{a}+2\right]}{\sqrt{a}-2}\))  \(.\dfrac{\sqrt{a}}{\sqrt{a}+2}\)

=\(\left(\sqrt{a}+2+\sqrt{a}+2\right).\dfrac{\sqrt{a}}{\sqrt{a}+2}\)

=\(\left(2\sqrt{a}+4\right).\dfrac{\sqrt{a}}{\sqrt{a}+2}\)

=\(2\left(\sqrt{a}+2\right).\dfrac{\sqrt{a}}{\sqrt{a}+2}\)

=\(2\sqrt{a}\)