Question

calculati media geometrica a numerelor x egal cu 2 radical din 18 plus 3 radical din 12 si y egal cu 12 supra radical din 8 plus radical din 12

Answer 1

[tex]x=2\sqrt{18}+3\sqrt{12}\\ x=2\cdot 3\sqrt{2}+3\cdot 2\sqrt3\\ x=6(\sqrt2+\sqrt3)\\ y=\frac{12}{\sqrt8+\sqrt12}\\ y=\frac{12}{2\sqrt2+2\sqrt3}\\ y=\frac{12}{2(\sqrt2+\sqrt3)}\\ y=\frac{6(\sqrt3-\sqrt2)}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}\\ y=6(\sqrt3-\sqrt2)\\ m_g=\sqrt{xy}\\ m_g=\sqrt{6(\sqrt2+\sqrt3)\cdot6(\sqrt3-\sqrt2)}\\ m_g=\sqrt{36(3-2)}\\ m_g=\boxed{\boxed{6}}[/tex]

Answer 2

$\it x=2\sqrt{18} +3\sqrt{12} = 2\cdot3(\sqrt2 + \sqrt3)$

$\it y = \dfrac{12}{\sqrt8+\sqrt{12}} = \dfrac{12}{2(\sqrt2+\sqrt3)}$

$\it m_g = \sqrt{x\cdot y} =\sqrt{ 2\cdot3(\sqrt2 + \sqrt3)\cdot\dfrac{12}{2(\sqrt2+\sqrt3)}} =\sqrt{36} =6\ .$