Question

cum se rezolva esercitiul;
media geometrica a numerelor;
a=[tex] \ \sqrt3+ \sqrt2
[/tex]
b=[tex]
\sqrt3- \sqrt2[/tex]

Answer 1

[tex]m_g=\sqrt{a*b}\\ m_g=\sqrt{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}\\ m_g=\sqrt{3-2}\\ m_g=1[/tex]

Answer 2

$m_{g}= \sqrt{a*b}$
$m_{g}= \sqrt{ \sqrt{3}+ \sqrt{2}* \sqrt{3}- \sqrt{2}}$
$m_{g}= \sqrt{ \sqrt{3} ^{2}- \sqrt{2} ^{2} }$
$m_{g}= \sqrt{3-2}= \sqrt{1}= 1$

Daca ai intrebari , sunt aici sa te ajut
Iti pot explica
 *hug*