Giải thích các bước giải:

a.Ta có:

$N=\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}$

$\to N=\dfrac{a}{\sqrt{ab}+b}-\dfrac{b}{a-\sqrt{ab}}-\dfrac{a+b}{\sqrt{ab}}$

$\to N=\dfrac{a}{\sqrt{b}(\sqrt{a}+\sqrt{b})}-\dfrac{b}{\sqrt{a}(\sqrt{a}-\sqrt{b})}-\dfrac{a+b}{\sqrt{ab}}$

$\to N=\dfrac{a\sqrt{a}(\sqrt{a}-\sqrt{b})-b\sqrt{b}(\sqrt{a}+\sqrt{b})-(a+b)(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}{\sqrt{ab}(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}$

$\to N=\dfrac{-\sqrt{ab}(a+b) }{\sqrt{ab}(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}$

$\to N=\dfrac{-(a+b) }{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}$

$\to N=\dfrac{-(a+b) }{a-b}$

 b.Ta có:

$a=\sqrt{6+2\sqrt5}=\sqrt{(\sqrt5+1)^2}=\sqrt5+1$

$b=\sqrt{6-2\sqrt5}=\sqrt{(\sqrt5-1)^2}=\sqrt5-1$

$\to a+b=2\sqrt5, a-b=2$

$\to N=\dfrac{-2\sqrt5}2=-\sqrt5$