Question

Ajutati ma va rog : (voi scrie radicalii cu/)
(/2+/3)^2+(/2-/3)^2=
(/2-/5)^2-(/2-/5)^2=
(/2/2-/6)^2+(4+/3)^2

Answer 1

Răspuns:

$({ \sqrt{2} + \sqrt{3} })^{2} + ({ \sqrt{2} - \sqrt{3} })^{2} = 2 + 2 \sqrt{6} + 3 + 2 - 2 \sqrt{6} + 3 = 10$

$( { \sqrt{2} - \sqrt{5} })^{2} - ( { \sqrt{2} - \sqrt{5} })^{2} = 2 - 2 \sqrt{10} + 5 - (2 - 2 \sqrt{10} + 5) = 7 - 2 \sqrt{10} - 7 + 2 \sqrt{10} = 0$

$( { \sqrt{2} \times \sqrt{2} - \sqrt{6} })^{2} + ({4 + \sqrt{3} })^{2} = ( {4 - \sqrt{6} })^{2} + ( {4} + \sqrt{3}) ^{2} = 16 - 8 \sqrt{6} + 6 + 16 + 8 \sqrt{3} + 3 = 46 - 8 \sqrt{6} + 8 \sqrt{3}$