Question

media geometrica a numerelor:a=(3-2radical din 2 )la a doua si b=(3+2 radical din 2) la a doua este ???

Answer 1

$a = {(3 - 2 \sqrt{2} )}^{2} = 9 - 12 \sqrt{2} + 8 = 17 - 12 \sqrt{2}$

$b = {(3 + 2 \sqrt{2} )}^{2} = 9 + 12 \sqrt{2} + 8 = 17 + 12 \sqrt{2}$

$m_{g} = \sqrt{ab}$

$m_{g} = \sqrt{(17 - 12 \sqrt{2} )(17 + 12 \sqrt{2} )}$

$m_{g} = \sqrt{ {17}^{2} - {(12 \sqrt{2}) }^{2} }$

$m_{g} = \sqrt{289 - 288}$

$m_{g} = \sqrt{1}$

$m_{g} = 1$