Question

cum se rezolva media geometrica a nr. a=√7-4√3 primul radical este peste toate numerele si 2+√3

Answer 1

    
$a= \sqrt{7-4 \sqrt{3} } = \\ =\sqrt{4+3-4 \sqrt{3} }= \\ =\sqrt{4-4 \sqrt{3}+3 }= \\ =\sqrt{ 2^{2} -2*2* \sqrt{3}+(-\sqrt{3})^2 }= \\ =\sqrt{ (2-\sqrt{3})^2 }=2-\sqrt{3} \\ Rezulta: \\ a = \boxed{2-\sqrt{3}} \\ \\ b=\boxed{2+\sqrt{3}} \\ \\ m_g= \sqrt{ab} =\sqrt{(2-\sqrt{3})(2+\sqrt{3})} =\sqrt{2^2-(\sqrt{3})^2} =\sqrt{4-3}= \sqrt{1} =\boxed{1}$