\(N=\left(2+\sqrt{a}\right)\left(\sqrt{a}-2\right)=a-4\)
\(N=\left(2+\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(\dfrac{\sqrt{a}+a}{1+a}-2\right)\)
\(N=\dfrac{2\sqrt{a}-6+a-3\sqrt{a}}{\sqrt{a}-3}.\left(\dfrac{\sqrt{a}\left(1+\sqrt{a}\right)}{1+\sqrt{a}}-2\right)\)
\(N=\dfrac{a-\sqrt{a}-6}{\sqrt{a}-3}.\left(\sqrt{a}-2\right)\)
\(N=\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-3}.\left(\sqrt{a}-2\right)\)
\(N=\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)\)
\(N=a-4\)
Với \(a\ge0,a\ne9\) ta có:
\(N=\left(2+\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(\dfrac{\sqrt{a}+a}{1+\sqrt{a}}-2\right)\)
\(=\dfrac{2\sqrt{a}-6+a-3\sqrt{a}}{\sqrt{a}-3}\cdot\dfrac{\sqrt{a}+a-2-2\sqrt{a}}{1+\sqrt{a}}\) \(=\dfrac{a-\sqrt{a}-6}{\sqrt{a}-3}\cdot\dfrac{a-\sqrt{a}-2}{1+\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-3}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\) \(=\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)\) \(=a-4\)
Vậy \(N=a-4\)
