Question

a=$\sqrt{3+2\sqrt{2}$+$\sqrt{3-2\sqrt{2}$

Aratati ca $a^{2}$ ∈ N

Answer 1

Răspuns:

8

Explicație pas cu pas:

Metoda 1 (prin depistarea patratelor perfecte):

$a=\sqrt{3+2\sqrt2}+\sqrt{3-2\sqrt2}\\a=\sqrt{(1+\sqrt2)^2}+\sqrt{(1-\sqrt2)^2}\\a=|1+\sqrt2|+|1-\sqrt2|\\a=1+\sqrt2+\sqrt2-1\\a=2\sqrt2\\a^2=8$

Metoda 2 (prin calcul direct):

$a=\sqrt{3+2\sqrt2}+\sqrt{3-2\sqrt2}\\a^2=(\sqrt{3+2\sqrt2}+\sqrt{3-2\sqrt2})^2\\a^2=3+2\sqrt2+3-2\sqrt2+2*\sqrt{3+2\sqrt2}*\sqrt{3-2\sqrt2}\\a^2=6+2*\sqrt{(3+2\sqrt2)*(3-2\sqrt2)}\\a^2=6+2*\sqrt{9-8}\\a^2=6+2\\a^2=8$