Question

media geometrica a numerelor x= radical 288si y=radical162 este?

Answer 1

x=√288=12√2
y=√162=9√2
Mg=√xy=√(12√2·9√2)=√(12·9·2)=√216=6√6

Answer 2

[tex]Mg= \sqrt{ \sqrt{288}* \sqrt{162} } [/tex]

= $\sqrt{12 \sqrt{2} * 9 \sqrt{2} }$

= $\sqrt{108 \sqrt{4} } = \sqrt{180*2} = \sqrt{216} = 6\sqrt{6}$