[tex]\it ._{a=\dfrac{\sqrt{24\cdot3}}{\sqrt{16\cdot2}-4(\sqrt2-3\sqrt2)}=\dfrac{\sqrt{72}}{4\sqrt2-4\sqrt2+12\sqrt2}=\dfrac{\sqrt{72}}{12\sqrt2}=\dfrac{\sqrt{36}}{12}=\dfrac{\ 6^{(6}}{12}=\dfrac{1}{2}}[/tex]
Pentru determinarea numărului b vom folosi formula:
[tex]\it \dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}\\ \\ \\ b=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}=1-\dfrac{1}{5}=\dfrac{4}{5}[/tex]
[tex]\it N=20(a+b)=20a+20b=20\cdot\dfrac{1}{2}+20\cdot\dfrac{4}{5}=10+16=26\\ \\ R\breve aspunsul\ corect\ este \ \ a)\ 26[/tex]