Question

media geometrica a numerelor
$a = |2 \sqrt{2} - 3|$
si
$b = |3 + 2\sqrt{2} |$

Answer 1

$2\sqrt{2}=\sqrt{8}<3=\sqrt{9} \implies |2\sqrt{2}-3|=3-2\sqrt{2}$

$3>0,\:\: 2\sqrt{2}>0\implies |3+2\sqrt{2}|=3+2\sqrt{2}$

$m_g=\sqrt{ab}=\sqrt{(3+2\sqrt{2})(3-2\sqrt{2})}=\sqrt{3^2-(2\sqrt{2})^2}=\sqrt{9-8}=1$

Answer 2

$\it 9>8 \Rightarrow \sqrt9>\sqrt8 \Rightarrow 3>\sqrt{4\cdot2} \Rightarrow 3>2\sqrt2 \Rightarrow 3-2\sqrt2>0 \Rightarrow \\ \\ \Rightarrow |3-2\sqrt2| =3-2\sqrt2 \ \ \ \ \ \ (*) \\ \\ a = |2\sqrt2-3| =|3-2\sqrt2| \stackrel{(*)}{=} 3-2\sqrt2 \\ \\ b= |3+2\sqrt2| =3+2\sqrt2$

$\it m_g=\sqrt{ab} = \sqrt{(3-2\sqrt2)(3+2\sqrt2)} = \sqrt{3^2-(2\sqrt2)^2} = \sqrt{9-8} =\sqrt1=1$