Va rog ,sa fie corect și sa dea că rezultat 1/2​

Răspuns:

[tex]a = \frac{ \sqrt{2}( \sqrt{2} - 1)}{2} + \frac{ \sqrt{6}( \sqrt{3} - \sqrt{2})}{6} + \frac{ \sqrt{3}(2 - \sqrt{3})}{6} = \frac{6 - 3 \sqrt{2} + 3 \sqrt{2} - 2 \sqrt{3} + 2 \sqrt{3} - 3}{6} = \frac{3}{6} = \frac{1}{2} [/tex]

[tex]\it \dfrac{\sqrt2-1}{\sqrt2}+\dfrac{\sqrt3-\sqrt2}{\sqrt6}+\dfrac{\sqrt4-\sqrt3}{\sqrt{12}}=\dfrac{\sqrt2}{\sqrt2}-\dfrac{1}{\sqrt2}+\dfrac{\sqrt3}{\sqrt6}-\dfrac{\sqrt2}{\sqrt6}+\dfrac{\sqrt4}{\sqrt{12}}-\dfrac{\sqrt3}{\sqrt{12}}=\\ \\ \\ =1-\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt2}-\dfrac{1}{\sqrt3}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{2}=1-\dfrac{1}{2}=\dfrac{1}{2}[/tex]