a) Fiecare dintre cele trei fracții este de forma :
[tex]\it \dfrac{a-b}{ab}= \dfrac{a}{ab}- \dfrac{b}{ab}= \dfrac{1}{b}- \dfrac{1}{a}\\ \\ \\ Deci:\\ \\ \\ \dfrac{\sqrt2-1}{\sqrt2}= \dfrac{1}{1}- \dfrac{1}{\sqrt2}=1-\dfrac{1}{\sqrt2}\\ \\ \\ \dfrac{\sqrt3-\sqrt2}{\sqrt6}= \dfrac{1}{\sqrt2}- \dfrac{1}{\sqrt3}\\ \\ \\ \dfrac{\sqrt4-\sqrt3}{\sqrt{12}}= \dfrac{1}{\sqrt3}- \dfrac{1}{\sqrt4}= \dfrac{1}{\sqrt3}- \dfrac{1}{2}[/tex]
[tex]\it a=1-\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt2}-\dfrac{1}{\sqrt3}+\dfrac{1}{\sqrt3}-\dfrac{1}{2}=1-\dfrac{1}{2}=\dfrac{1}{2}[/tex]
b)
[tex]\it \sqrt{n\cdot a}=\sqrt{n\cdot\dfrac{1}{2}}=\sqrt{\dfrac{n}{2}} \in \mathbb{N} \Rightarrow n=2k^2\\ \\ \\ n<100 \Rightarrow 2k^2<100|_{:2} \Rightarrow k^2<50 \Rightarrow k^2\in\{0,\ 1,\ 4,\ 9,\ 16,\ 25,\ 36,\ 49\}|_{\cdot2}\Rightarrow\\ \\ n=2k^2\in\{0,\ 2,\ 8,\ 18,\ 32,\ 50,\ 72,\ 98\}[/tex]
