Question

√3 supra 2-√3 + 3(√2-2) supra √3

Answer 1

Răspuns:

rad3(2 + rad3)/(4-3) + 3rad3(rad2 - 2)/3 =

2rad3 + 3 + rad6 - 2rad3 =

3 + rad6

Asta, dacă numitorul este (2-rad3)

Answer 2

$\it \dfrac{\sqrt3}{2}-\sqrt3+\dfrac{^{\sqrt3)}3(\sqrt2-2)}{\sqrt3}=\dfrac{\sqrt3}{2}-\sqrt3+\dfrac{3\sqrt3(\sqrt2-2)}{3}=\dfrac{\sqrt3}{2}-\sqrt3+\sqrt6-2\sqrt3\\ \\ =\ ^{2)}\sqrt6-\ ^{2)}3\sqrt3+\dfrac{\sqrt3}{2}=\dfrac{2\sqrt6-6\sqrt3+\sqrt3}{2}=\dfrac{2\sqrt3-5\sqrt3}{2}$