Răspuns:
rad3(2 + rad3)/(4-3) + 3rad3(rad2 - 2)/3 =
2rad3 + 3 + rad6 - 2rad3 =
3 + rad6
Asta, dacă numitorul este (2-rad3)
[tex]\it \dfrac{\sqrt3}{2}-\sqrt3+\dfrac{^{\sqrt3)}3(\sqrt2-2)}{\sqrt3}=\dfrac{\sqrt3}{2}-\sqrt3+\dfrac{3\sqrt3(\sqrt2-2)}{3}=\dfrac{\sqrt3}{2}-\sqrt3+\sqrt6-2\sqrt3\\ \\ =\ ^{2)}\sqrt6-\ ^{2)}3\sqrt3+\dfrac{\sqrt3}{2}=\dfrac{2\sqrt6-6\sqrt3+\sqrt3}{2}=\dfrac{2\sqrt3-5\sqrt3}{2}[/tex]
