Question

Media geometrica va roooog

Answer 1

$\it a = \dfrac{1}{2-\sqrt3} +\sqrt{7+4\sqrt3}$

$\it \dfrac{1}{2-\sqrt3} = \dfrac{1}{2-\sqrt3} \cdot \dfrac{2+\sqrt3}{2+\sqrt3}= \dfrac{2+\sqrt3}{4-3} = 2+\sqrt3$

[tex]\it \sqrt{7+4\sqrt3} = \sqrt{4+3+4\sqrt3} = \sqrt{2^2+4\sqrt3+(\sqrt3)^2}= \\ \\ =\sqrt{(2+\sqrt3)^2} =2+\sqrt3[/tex]

$\it a= 2+\sqrt3+2+\sqrt3 =4+2\sqrt3$

[tex]\it b = 2-\sqrt3+\dfrac{1}{2+\sqrt3} = 2-\sqrt3+\dfrac{2-\sqrt3}{4- 3}= 2-\sqrt3+2-\sqrt3 \\ \\ = 4-2\sqrt3[/tex]

[tex]\it m_g = \sqrt{\it a\cdot b} =\sqrt{(4+2\sqrt3)(4-2\sqrt3)} =\sqrt{4^2-(2\sqrt3)^2}= \\ \\ \sqrt{16-12} =\sqrt4 =2[/tex]