Question

Media geometrica a numerelor : (2radical3 +1)la puterea a doua , 13-4radical3

Answer 1

Media geometrica = $\bf \sqrt{a\cdot b}$ , a,b ≥ 0

Numerele: $\bf (2\sqrt{3}+1)^{2}$   si    $\bf 13-4\sqrt{3}$

$\bf Mg = \sqrt{(2\sqrt{3}+1)^{2}\cdot(13-4\sqrt{3} )}$

$\bf Mg = \sqrt{(2\sqrt{3})^{2}+2\cdot2\sqrt{3} +1)\cdot(13-4\sqrt{3} )}$

$\bf Mg = \sqrt{(4\cdot3+4\sqrt{3} +1)\cdot(13-4\sqrt{3} )}$

$\bf Mg = \sqrt{(12+4\sqrt{3} +1)\cdot(13-4\sqrt{3} )}$

$\bf Mg = \sqrt{(13+4\sqrt{3})\cdot(13-4\sqrt{3} )}$

$\bf Mg = \sqrt{13^{2}-(4\sqrt{3})^{2}}$

$\bf Mg = \sqrt{169-4^{2}\cdot(\sqrt{3})^{2}}$

$\bf Mg = \sqrt{169-16\cdot3}}$

$\bf Mg = \sqrt{169-48}$

$\bf Mg = \sqrt{121}$

$\bf Mg = \sqrt{11^{2}}$

$\boxed{\bf Mg = 11}$

Raspuns: 11 este media geometrica a numerelor