Question

(1+radical 2) 2+(1-radical 2)2

Answer 1

   
[tex]\displaystyle\\ \Big(1 + \sqrt{2} \Big)^2+\Big(1 - \sqrt{2} \Big)^2 = \\\\ =\left(1^2 + 2\times 1\times \sqrt{2} + \Big(\sqrt{2}\Big)^2\right)+\left(1^2 + 2\times 1\times (-\sqrt{2}) + \Big(-\sqrt{2}\Big)^2\right)=\\\\ 1+\underline{2\sqrt{2}} +2 + 1-\underline{2\sqrt{2}} +2 = 1 +2+1+2 = \boxed{\bf 6} [/tex]

Answer 2

Pentru rezolvarea acestui exercitiu trebuie sa ai in vedere , cunoașterea formulelelor de calcul prescurtat.
Si anume : (a+b)^2 = a^2+2ab+b^2
(a-b)^2= a^2 - 2ab + b^2

( 1+ rad2)^2 = 1+2rad2+2
( 1- rad2)^2 = 1-2rad2 +2
Acum rescriem ecuatia
1+2rad2+2 +1-2rad2+2
Scadem termeni opusi : 2rad2 si (-2rad2)
Rămâne : 1+2+1+2 = 3+3= 6